Chemistry Quiz: Interpret Nuclear Equations
20 questions · exam conditions
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Interpret Nuclear EquationsQuestion 1 of 20

Which product nucleus forms in the beta decay 614C→(product)+−10e?^{14}_{6}\text{C} \rightarrow \text{(product)} + ^{0}_{-1}\text{e}?

514B^{14}_{5}\text{B}
713N^{13}_{7}\text{N}
714N^{14}_{7}\text{N}
410Be^{10}_{4}\text{Be}
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Chemistry Quiz

Chemistry Quiz: Interpret Nuclear Equations

Practice Interpret Nuclear Equations in Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Interpret Nuclear Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which product nucleus forms in the beta decay 614C→(product)+−10e?^{14}_{6}\text{C} \rightarrow \text{(product)} + ^{0}_{-1}\text{e}?

  1. 514B^{14}_{5}\text{B}
  2. 713N^{13}_{7}\text{N}
  3. 714N^{14}_{7}\text{N} (correct answer)
  4. 410Be^{10}_{4}\text{Be}
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 92238U^{238}_{92}\text{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 92238U→90234Th+24He^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (24He^{4}_{2}\text{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (24He^{4}_{2}\text{He}), beta (−10e^{0}_{-1}\text{e}, electron), neutron (01n^{1}_{0}\text{n}), proton (11H^{1}_{1}\text{H}). In beta decay of 614C→^{14}_{6}\text{C} \rightarrow product + −10e^{0}_{-1}\text{e}, mass stays 14 (14=14+0) and atomic increases to 7 (6=7-1), so the product is 714N^{14}_{7}\text{N}. Choice C correctly interprets the nuclear equation by identifying the product nucleus as 714N^{14}_{7}\text{N}. Choice B (713N^{13}_{7}\text{N}) fails as a distractor because it has mass 13, but beta decay doesn't change mass number. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 88226Ra→86222Rn+X^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + \text{X}. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\text{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (24He^{4}_{2}\text{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\text{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (01n^{1}_{0}\text{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. Awesome job on beta decay!

Question 2

In the fusion equation 12H+12H→23He+X^{2}_{1}\text{H} + ^{2}_{1}\text{H} \rightarrow ^{3}_{2}\text{He} + X, which particle must XX be to conserve mass number and atomic number?

  1. 11H^{1}_{1}\text{H} (proton)
  2. 01n^{1}_{0}\text{n} (neutron) (correct answer)
  3. −10e^{0}_{-1}\text{e} (beta particle)
  4. 24He^{4}_{2}\text{He} (alpha particle)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 92238U^{238}_{92}\text{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 92238U→90234Th+24He^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (24He^{4}_{2}\text{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (24He^{4}_{2}\text{He}), beta (−10e^{0}_{-1}\text{e}, electron), neutron (01n^{1}_{0}\text{n}), proton (11H^{1}_{1}\text{H}). For fusion 12H+12H→23He+X^{2}_{1}\text{H} + ^{2}_{1}\text{H} \rightarrow ^{3}_{2}\text{He} + X, masses balance as 2+2=3+1 and atomics as 1+1=2+0, so X must be 01n^{1}_{0}\text{n}, a neutron. Choice B correctly interprets the nuclear equation by identifying X as a neutron to conserve both mass and atomic numbers. Choice D (alpha particle) fails because an alpha has mass 4 and atomic 2, which would unbalance the equation (left mass 4, right 3+4=7). Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 88226Ra→86222Rn+X^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + X. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\text{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (24He^{4}_{2}\text{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\text{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (01n^{1}_{0}\text{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. You're a fusion expert now!

Question 3

In the nuclear equation 92238U→90234Th+X^{238}_{92}\mathrm{U} \rightarrow ^{234}_{90}\mathrm{Th} + X, what particle is XX?

  1. −10e^{0}_{-1}\mathrm{e} (beta particle)
  2. 24He^{4}_{2}\mathrm{He} (alpha particle) (correct answer)
  3. 11H^{1}_{1}\mathrm{H} (proton)
  4. 01n^{1}_{0}\mathrm{n} (neutron)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 92238U^{238}_{92} \mathrm{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 92238U→90234Th+24He^{238}_{92} \mathrm{U} \rightarrow ^{234}_{90} \mathrm{Th} + ^{4}_{2} \mathrm{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle ( 24He^{4}_{2} \mathrm{He} , which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (24He^{4}_{2} \mathrm{He}), beta (−10e^{0}_{-1} \mathrm{e}, electron), neutron (01n^{1}_{0} \mathrm{n}), proton (11H^{1}_{1} \mathrm{H}). In this equation, 92238U→90234Th+X^{238}_{92} \mathrm{U} \rightarrow ^{234}_{90} \mathrm{Th} + X, to find X, subtract the mass and atomic numbers: mass 238 - 234 = 4, atomic 92 - 90 = 2, so X is 24He^{4}_{2} \mathrm{He}, an alpha particle. Choice B correctly interprets the nuclear equation by properly identifying X as the alpha particle based on conservation laws. A common distractor like choice A might confuse this with beta decay, but beta would change atomic number by +1 without mass change, not matching here—remember, alpha reduces mass by 4 and atomic by 2. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 88226Ra→86222Rn+X^{226}_{88} \mathrm{Ra} \rightarrow ^{222}_{86} \mathrm{Rn} + X. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2} \mathrm{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (24He^{4}_{2} \mathrm{He}). If atomic number increases by 1 with no mass change → beta emitted (−10e^{0}_{-1} \mathrm{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (01n^{1}_{0} \mathrm{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. Keep practicing, and you'll master these equations effortlessly!

Question 4

In beta decay, a nucleus emits an electron (−10e^{0}_{-1}\text{e}). If ZAX→Z+1AY+−10e,^{A}_{Z}\text{X} \rightarrow ^{A}_{Z+1}\text{Y} + ^{0}_{-1}\text{e}, what happens to the mass number AA?

  1. It increases by 1.
  2. It decreases by 1.
  3. It stays the same. (correct answer)
  4. It decreases by 4.
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 238U^{238}\text{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 238U→234Th+4He^{238}\text{U} \rightarrow ^{234}\text{Th} + ^4\text{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (4He^4\text{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (4He^4\text{He}), beta (−10e^{0}_{-1}\text{e}, electron), neutron (1n^1\text{n}), proton (1H^1\text{H}). In beta decay like ZAX→Z+1AY+−10e^{A}_{Z}\text{X} \rightarrow ^{A}_{Z+1}\text{Y} + ^{0}_{-1}\text{e}, the mass number A remains unchanged because the emitted beta particle (electron) has negligible mass (0), so A = A + 0. Choice C correctly interprets the nuclear equation by stating that the mass number stays the same. Choice D (decreases by 4) fails as a distractor because that describes alpha decay, not beta, where mass drops by 4. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 226Ra→222Rn+X^{226}\text{Ra} \rightarrow ^{222}\text{Rn} + \text{X}. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^4_2\text{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (4He^4\text{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\text{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (1n^1\text{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. Keep up the fantastic progress!

Question 5

Deuterium-deuterium reactions can be written as 12H+12H→23He+01n.^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \rightarrow {}^{3}_{2}\text{He} + {}^{1}_{0}n. What process is shown by this equation?

  1. Fusion (correct answer)
  2. Fission
  3. Alpha decay
  4. Beta decay
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): ²H (deuterium) has mass 2 and atomic 1. The equation ²₁H + ²₁H → ³₂He + ¹₀n shows two deuterium nuclei combining to form helium-3 plus a neutron, with conservation check: mass (2+2 = 3+1 ✓) and atomic (1+1 = 2+0 ✓)—this is fusion where light nuclei combine. Choice A correctly identifies this as fusion, recognizing the characteristic pattern of two light nuclei (hydrogen isotopes) combining into a heavier nucleus (helium) with particle emission. Choice B (fission) splits heavy nuclei, choice C (alpha decay) emits helium-4, and choice D (beta decay) changes one nucleus—none match this combining pattern. Fusion identification keys: (1) Light nuclei on left (usually H, He, Li), (2) Heavier nucleus on right (but still light elements), (3) Often releases neutrons or protons, (4) Powers stars including our sun! D-D fusion (shown here) is one pathway in stellar nucleosynthesis. The key difference from fission: fusion COMBINES light nuclei (releases energy), fission SPLITS heavy nuclei (also releases energy), but fusion produces no radioactive waste!

Question 6

A fusion reaction is shown: 12H+13H→24He+01n.^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \rightarrow {}^{4}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}. What particle is produced along with helium-4?

  1. 11H^{1}_{1}\mathrm{H} (proton)
  2. −10e^{0}_{-1}\mathrm{e} (beta particle)
  3. 01n^{1}_{0}\mathrm{n} (neutron) (correct answer)
  4. 24He^{4}_{2}\mathrm{He} (alpha particle)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 92238U^{238}_{92}\mathrm{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 92238U→90234Th+24He^{238}_{92}\mathrm{U} \rightarrow ^{234}_{90}\mathrm{Th} + ^{4}_{2}\mathrm{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle ( 24He^{4}_{2}\mathrm{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha ( 24He^{4}_{2}\mathrm{He}), beta ( −10e^{0}_{-1}\mathrm{e}, electron), neutron ( 01n^{1}_{0}\mathrm{n}), proton ( 11H^{1}_{1}\mathrm{H}). In this fusion equation, 12H+13H→24He+01n^{2}_{1}\mathrm{H} + ^{3}_{1}\mathrm{H} \rightarrow ^{4}_{2}\mathrm{He} + ^{1}_{0}\mathrm{n}, two light hydrogen isotopes combine to form helium-4 and a neutron, with mass (5=4+1) and atomic (2=2+0) conserved. Choice C correctly interprets the nuclear equation by properly identifying the particle produced as a neutron. A distractor like choice D (alpha particle) fails because alpha is 24He^{4}_{2}\mathrm{He}, which is already a product, not the additional particle—double-check all products. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 88226Ra→86222Rn+X^{226}_{88}\mathrm{Ra} \rightarrow ^{222}_{86}\mathrm{Rn} + \mathrm{X}. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\mathrm{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted ( 24He^{4}_{2}\mathrm{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\mathrm{e}). If just gamma (energy), no mass or atomic change. If neutrons involved ( 01n^{1}_{0}\mathrm{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. You're building strong skills—keep it up!

Question 7

In the nuclear equation 84210Po→ 82206Pb+X,^{210}_{84}\text{Po} \rightarrow \,^{206}_{82}\text{Pb} + X, what is XX?

  1. 24He^{4}_{2}\text{He} (alpha particle) (correct answer)
  2. −10e^{0}_{-1}\text{e} (beta particle)
  3. 01n^{1}_{0}\text{n} (neutron)
  4. 00γ^{0}_{0}\gamma (gamma ray)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 238U^{238}\text{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 238U→234Th+4He^{238}\text{U} \rightarrow ^{234}\text{Th} + ^{4}\text{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (4He^{4}\text{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (4He^{4}\text{He}), beta (−10e^{0}_{-1}\text{e}, electron), neutron (1n^{1}\text{n}), proton (1H^{1}\text{H}). For 84210Po→82206Pb+X^{210}_{84}\text{Po} \rightarrow ^{206}_{82}\text{Pb} + X, mass balances as 210=206+4 and atomic as 84=82+2, so X is 24He^{4}_{2}\text{He}, an alpha particle. Choice A correctly interprets the nuclear equation by identifying X as an alpha particle. Choice B (beta particle) fails because a beta would keep mass at 210 and change atomic to 85, not matching the decrease to 82. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 226Ra→222Rn+X^{226}\text{Ra} \rightarrow ^{222}\text{Rn} + X. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\text{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (4He^{4}\text{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\text{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (1n^{1}\text{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. You're excelling at this!

Question 8

A nucleus undergoes beta decay: 53131I→X+−10e.^{131}_{53}\text{I} \rightarrow X + {}^{0}_{-1}e. What is the product nucleus XX?

  1. 52131Te^{131}_{52}\text{Te}
  2. 53127I^{127}_{53}\text{I}
  3. 54131Xe^{131}_{54}\text{Xe} (correct answer)
  4. 55135Cs^{135}_{55}\text{Cs}
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): iodine-131 has mass 131 and atomic 53. In beta decay ¹³¹₅₃I → X + ⁰₋₁e, a beta particle (electron) with mass 0 and atomic number -1 is emitted, so using conservation: mass stays 131 (131 = X + 0) and atomic number becomes 54 (53 = X + (-1)), making X = ¹³¹₅₄Xe. Choice C correctly identifies the product as ¹³¹₅₄Xe (xenon-131), recognizing that beta decay increases atomic number by 1 (I→Xe, 53→54) while preserving mass number. Choice A has wrong atomic number (52), choice B has wrong mass (127), and choice D has both wrong—only xenon-131 satisfies conservation. Beta decay pattern recognition: (1) Mass number NEVER changes in beta decay, (2) Atomic number ALWAYS increases by 1, (3) Element changes to next one on periodic table (I→Xe, C→N, etc.). This happens because a neutron converts to a proton inside the nucleus, emitting an electron to conserve charge. Memorizing this +1 atomic rule makes beta decay problems instant to solve!

Question 9

In the alpha decay equation 92238U→90234Th+24He,^{238}_{92}\mathrm{U} \rightarrow {}^{234}_{90}\mathrm{Th} + {}^{4}_{2}\mathrm{He}, what happens to the atomic number of the nucleus after the decay?

  1. It increases by 2.
  2. It decreases by 2. (correct answer)
  3. It stays the same.
  4. It decreases by 4.
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, ²³⁸U represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: ²³⁸U → ²³⁴Th + ⁴He means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (⁴He, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (⁴He), beta (⁰₋₁e, electron), neutron (¹n), proton (¹H). In this alpha decay, ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He, the atomic number drops from 92 to 90, decreasing by 2, as alpha emission removes 2 protons. Choice B correctly interprets the nuclear equation by noting the atomic number decreases by 2. A distractor like choice D fails because it confuses atomic number change (dec by 2) with mass number change (dec by 4)—focus on which number you're tracking. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: ²²⁶Ra → ²²²Rn + X. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is ⁴₂He (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (⁴He). If atomic number increases by 1 with no mass change → beta emitted (electron, ⁰₋₁e). If just gamma (energy), no mass or atomic change. If neutrons involved (¹n) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. Impressive attention to changes—continue strong!

Question 10

A nucleus emits a beta particle: 1532P→1632S+−10e.^{32}_{15}\mathrm{P} \rightarrow {}^{32}_{16}\mathrm{S} + {}^{0}_{-1}\mathrm{e}. Which quantity stays the same in this reaction?

  1. Mass number of the nucleus (correct answer)
  2. Atomic number of the nucleus
  3. Element identity (symbol) of the nucleus
  4. Number of protons in the nucleus
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 238U^{238}\mathrm{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 238U→234Th+4He^{238}\mathrm{U} \rightarrow ^{234}\mathrm{Th} + ^{4}\mathrm{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (4He^{4}\mathrm{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (4He^{4}\mathrm{He}), beta (−10e^{0}_{-1}\mathrm{e}, electron), neutron (1n^{1}\mathrm{n}), proton (1H^{1}\mathrm{H}). In this beta decay, 1532P→1632S+−10e^{32}_{15}\mathrm{P} \rightarrow ^{32}_{16}\mathrm{S} + ^{0}_{-1}\mathrm{e}, the mass number remains 32 (32=32+0), while atomic number increases from 15 to 16, changing the element but keeping mass the same. Choice A correctly interprets the nuclear equation by identifying that the mass number of the nucleus stays the same. A distractor like choice B fails because atomic number changes (increases by 1 in beta decay)—recall that mass is conserved, but protons adjust. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 226Ra→222Rn+X^{226}\mathrm{Ra} \rightarrow ^{222}\mathrm{Rn} + \mathrm{X}. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\mathrm{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (4He^{4}\mathrm{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\mathrm{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (1n^{1}\mathrm{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. You're mastering conservation—bravo!

Question 11

In the nuclear equation 92238U→90234Th+X^{238}_{92}\mathrm{U} \rightarrow {}^{234}_{90}\mathrm{Th} + X, what particle is represented by XX?

  1. −10e^{0}_{-1}\mathrm{e} (beta particle)
  2. 01n^{1}_{0}\mathrm{n} (neutron)
  3. 24He^{4}_{2}\mathrm{He} (alpha particle) (correct answer)
  4. 11H^{1}_{1}\mathrm{H} (proton)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 92238U^{238}_{92}\mathrm{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 92238U→90234Th+24He^{238}_{92}\mathrm{U} \rightarrow ^{234}_{90}\mathrm{Th} + ^{4}_{2}\mathrm{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (24He^{4}_{2}\mathrm{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (24He^{4}_{2}\mathrm{He}), beta (−10e^{0}_{-1}\mathrm{e}, electron), neutron (01n^{1}_{0}\mathrm{n}), proton (11H^{1}_{1}\mathrm{H}). In this equation, 92238U→90234Th+X^{238}_{92}\mathrm{U} \rightarrow ^{234}_{90}\mathrm{Th} + X, to find X, check conservation: mass 238 = 234 + A so A=4, atomic 92=90+Z so Z=2, making X an alpha particle (24He^{4}_{2}\mathrm{He}). Choice C correctly interprets the nuclear equation by properly identifying the particle as an alpha particle based on the conservation of mass and atomic numbers. A common distractor like choice A (beta particle) fails because a beta would have mass 0 and atomic -1, which wouldn't balance the mass drop of 4 or atomic drop of 2—remember, betas keep mass the same but increase atomic number by 1. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 88226Ra→86222Rn+X^{226}_{88}\mathrm{Ra} \rightarrow ^{222}_{86}\mathrm{Rn} + X. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\mathrm{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (24He^{4}_{2}\mathrm{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\mathrm{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (01n^{1}_{0}\mathrm{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. Keep practicing, and you'll master these balances effortlessly!

Question 12

In the fission equation 92235U+01n→56144Ba+3689Kr+3 01n^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{144}_{56}\mathrm{Ba} + {}^{89}_{36}\mathrm{Kr} + 3\,{}^{1}_{0}\mathrm{n}, which process is represented?

  1. Alpha decay
  2. Beta decay
  3. Nuclear fission (correct answer)
  4. Nuclear fusion
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 238U^{238}\mathrm{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 238U→234Th+4He^{238}\mathrm{U} \rightarrow ^{234}\mathrm{Th} + ^{4}\mathrm{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle ( 4He^{4}\mathrm{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238=234+4238 = 234 + 4 ✓ and 92=90+292 = 90 + 2 ✓. Common particles: alpha ( 4He^{4}\mathrm{He}), beta ( −10e^{0}_{-1}\mathrm{e}, electron), neutron ( 1n^{1}\mathrm{n}), proton ( 1H^{1}\mathrm{H}). This equation shows 92235U+01n→56144Ba+3689Kr+3 01n^{235}_{92}\mathrm{U} + ^{1}_{0}\mathrm{n} \rightarrow ^{144}_{56}\mathrm{Ba} + ^{89}_{36}\mathrm{Kr} + 3\, ^{1}_{0}\mathrm{n}, where a heavy uranium nucleus absorbs a neutron and splits into two medium-sized nuclei plus more neutrons, with mass ( 236=144+89+3236=144+89+3 ) and atomic ( 92=56+3692=56+36 ) conserved, typical of fission. Choice C correctly interprets the nuclear equation by recognizing the nuclear process type as fission due to the splitting of a heavy nucleus. A distractor like choice D (fusion) fails because fusion involves light nuclei combining, not splitting—look for combining vs. splitting patterns. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 226Ra→222Rn+X^{226}\mathrm{Ra} \rightarrow ^{222}\mathrm{Rn} + \mathrm{X}. Mass: 226=222+?226 = 222 + ? → X has mass 4. Atomic: 88=86+?88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\mathrm{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted ( 4He^{4}\mathrm{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\mathrm{e}). If just gamma (energy), no mass or atomic change. If neutrons involved ( 1n^{1}\mathrm{n} ) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients ( 2H2+O2→2H2O2\mathrm{H}_{2} + \mathrm{O}_{2} \rightarrow 2\mathrm{H}_{2}\mathrm{O} keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. Great job identifying processes—keep going!

Question 13

In the nuclear equation 84210Po→82206Pb+X,^{210}_{84}\mathrm{Po} \rightarrow {}^{206}_{82}\mathrm{Pb} + X, what is XX?

  1. 00γ^{0}_{0}\gamma (gamma ray)
  2. −10e^{0}_{-1}\mathrm{e} (beta particle)
  3. 24He^{4}_{2}\mathrm{He} (alpha particle) (correct answer)
  4. 01n^{1}_{0}\mathrm{n} (neutron)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 238U^{238}\mathrm{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 238U→234Th+4He^{238}\mathrm{U} \rightarrow ^{234}\mathrm{Th} + ^{4}\mathrm{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (24He^{4}_{2}\mathrm{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238=234+4238 = 234 + 4 ✓ and 92=90+292 = 90 + 2 ✓. Common particles: alpha (24He^{4}_{2}\mathrm{He}), beta (−10e^{0}_{-1}\mathrm{e}, electron), neutron (01n^{1}_{0}\mathrm{n}), proton (11H^{1}_{1}\mathrm{H}). For 84210Po→82206Pb+X^{210}_{84}\mathrm{Po} \rightarrow ^{206}_{82}\mathrm{Pb} + X, conservation gives mass 210=206+A210 = 206 + A so A=4A = 4, atomic 84=82+Z84 = 82 + Z so Z=2Z = 2, identifying X as an alpha particle (24He^{4}_{2}\mathrm{He}). Choice C correctly interprets the nuclear equation by properly identifying the particle as an alpha particle. A distractor like choice B (beta particle) fails because it would require no mass change and an atomic increase, but here mass drops by 4—verify both mass and atomic balances. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 88226Ra→86222Rn+X^{226}_{88}\mathrm{Ra} \rightarrow ^{222}_{86}\mathrm{Rn} + X. Mass: 226=222+?226 = 222 + ? → X has mass 4. Atomic: 88=86+?88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\mathrm{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (24He^{4}_{2}\mathrm{He}). If atomic number increases by 1 with no mass change → beta emitted (−10e^{0}_{-1}\mathrm{e}). If just gamma (00γ^{0}_{0}\gamma), no mass or atomic change. If neutrons involved (01n^{1}_{0}\mathrm{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. You're nailing particle identification—keep practicing!

Question 14

A decay is shown: 1940K→2040Ca+X.^{40}_{19}\text{K} \rightarrow ^{40}_{20}\text{Ca} + X. Which particle is XX?

  1. 24He^{4}_{2}\text{He} (alpha particle)
  2. 01n^{1}_{0}\text{n} (neutron)
  3. −10e^{0}_{-1}\text{e} (beta particle) (correct answer)
  4. 11H^{1}_{1}\text{H} (proton)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): potassium-40 is ⁴⁰₁₉K. The equation ⁴⁰₁₉K → ⁴⁰₂₀Ca + X shows potassium-40 becoming calcium-40, where mass stays 40 but atomic number increases from 19 to 20—this is the signature of beta decay. Using conservation: X must have mass 0 (since 40 = 40 + 0) and atomic number -1 (since 19 = 20 + (-1)), which matches a beta particle. Choice C correctly identifies X as ⁰₋₁e (beta particle) because it has mass 0 and charge -1, satisfying both conservation requirements. The other particles don't work: alpha has mass 4, neutron has no charge, proton has positive charge—only beta particle conserves both quantities. Beta decay recognition in this problem: (1) Mass number unchanged (40 → 40), (2) Atomic number increases by 1 (19 → 20), (3) This means a neutron converted to proton, emitting an electron. Potassium-40 is naturally radioactive and found in bananas—it's one source of natural background radiation we're all exposed to daily!

Question 15

Radium-226 decays according to 88226Ra→86222Rn+24He.^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + ^{4}_{2}\text{He}. Which particle is emitted?

  1. A neutron (01n^{1}_{0}\text{n})
  2. A gamma ray (γ\gamma)
  3. A beta particle (−10e^{0}_{-1}\text{e})
  4. An alpha particle (24He^{4}_{2}\text{He}) (correct answer)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): radium-226 is ²²⁶₈₈Ra with mass 226 and atomic number 88. The equation ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He clearly shows radium-226 decaying to radon-222 plus a particle with mass 4 and atomic number 2, which is an alpha particle (helium nucleus). Choice D correctly identifies the emitted particle as an alpha particle (⁴₂He) because it's explicitly written in the equation with mass number 4 and atomic number 2. The other particles have different signatures: neutron (¹₀n), gamma ray (no mass or charge), beta particle (⁰₋₁e)—none match what's shown in the equation. Reading emission identification: (1) Look at what's written after the arrow on the product side, (2) Match the mass and atomic numbers to known particles, (3) Verify conservation: 226 = 222 + 4 ✓ and 88 = 86 + 2 ✓. Alpha emission is common for heavy radioactive elements like radium because alpha particles are stable and effectively reduce both mass and atomic number, moving the nucleus toward stability!

Question 16

In the fusion reaction 12H+13H→24He+01n,^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He} + {}^{1}_{0}n, which particle is produced along with helium-4?

  1. 11H^{1}_{1}\text{H} (proton)
  2. −10e^{0}_{-1}e (beta particle)
  3. 01n^{1}_{0}n (neutron) (correct answer)
  4. 24He^{4}_{2}\text{He} (alpha particle)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, ²H represents deuterium (heavy hydrogen) with mass number 2 and atomic number 1. The fusion equation ²₁H + ³₁H → ⁴₂He + ¹₀n shows deuterium and tritium combining to form helium-4 plus a neutron, where we can verify conservation: mass (2+3 = 4+1 ✓) and atomic number (1+1 = 2+0 ✓). Choice C correctly identifies the neutron (¹₀n) as the particle produced alongside helium-4, with its characteristic mass number 1 and atomic number 0. Choice A (proton) would have atomic number 1, making the equation unbalanced, while choices B and D would violate mass or charge conservation. Fusion reactions combine light nuclei into heavier ones, releasing enormous energy—this powers the sun! Common fusion patterns: (1) D-T fusion (shown here) produces He-4 + neutron, (2) D-D fusion can produce He-3 + neutron or tritium + proton, (3) Multiple hydrogen isotopes fusing always produce helium plus particles. The key to reading fusion equations is checking that small nuclei on the left combine into larger nucleus plus particles on the right, with perfect conservation of mass and atomic numbers.

Question 17

A simplified fission reaction is shown: 92235U+01n→56144Ba+3689Kr+301n.^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{144}_{56}\text{Ba} + ^{89}_{36}\text{Kr} + 3^{1}_{0}\text{n}. Which statement best describes this nuclear process?

  1. Two light nuclei combine to form a heavier nucleus (fusion).
  2. A heavy nucleus splits into smaller nuclei and releases neutrons (fission). (correct answer)
  3. A nucleus emits an alpha particle and becomes a new element (alpha decay).
  4. A nucleus emits an electron, decreasing its atomic number by 1 (beta decay).
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 92238U^{238}_{92}\text{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 92238U→90234Th+24He^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle ( 24He^{4}_{2}\text{He} , which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha ( 24He^{4}_{2}\text{He} ), beta ( −10e^{0}_{-1}\text{e} , electron), neutron ( 01n^{1}_{0}\text{n} ), proton ( 11H^{1}_{1}\text{H} ). This equation shows 92235U+01n→56144Ba+3689Kr+301n^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{144}_{56}\text{Ba} + ^{89}_{36}\text{Kr} + 3^{1}_{0}\text{n}, where a heavy nucleus absorbs a neutron and splits into two smaller nuclei while releasing more neutrons, which is fission. Choice B correctly interprets the nuclear equation by describing it as a heavy nucleus splitting into smaller nuclei and releasing neutrons (fission). Choice A fails as a distractor because it describes fusion, where light nuclei combine, but here a heavy nucleus is splitting, not combining. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 88226Ra→86222Rn+X^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + \text{X}. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\text{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (24He^{4}_{2}\text{He}). If atomic number increases by 1 with no mass change → beta emitted (−10e^{0}_{-1}\text{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (01n^{1}_{0}\text{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. Excellent work distinguishing fission!

Question 18

In the nuclear equation 94238Pu→ 92234U+ 24He,^{238}_{94}\mathrm{Pu} \rightarrow \,^{234}_{92}\mathrm{U} + \,^{4}_{2}\mathrm{He}, what is the product nucleus formed (the element and mass number)?

  1. Uranium-234 (92234U^{234}_{92}\mathrm{U}) (correct answer)
  2. Uranium-238 (92238U^{238}_{92}\mathrm{U})
  3. Thorium-234 (90234Th^{234}_{90}\mathrm{Th})
  4. Neptunium-234 (93234Np^{234}_{93}\mathrm{Np})
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 238U^{238}\mathrm{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 92238U→90234Th+24He^{238}_{92}\mathrm{U} \rightarrow ^{234}_{90}\mathrm{Th} + ^{4}_{2}\mathrm{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (24He^{4}_{2}\mathrm{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (24He^{4}_{2}\mathrm{He}), beta (−10e^{0}_{-1}\mathrm{e}, electron), neutron (1n^{1}\mathrm{n}), proton (11H^{1}_{1}\mathrm{H}). The equation 94238Pu→92234U+24He^{238}_{94}\mathrm{Pu} \rightarrow ^{234}_{92}\mathrm{U} + ^{4}_{2}\mathrm{He} shows plutonium decaying via alpha emission to uranium-234, with mass (238=234+4) and atomic (94=92+2) balanced. Choice A correctly interprets the nuclear equation by identifying the product nucleus as uranium-234 (92234U^{234}_{92}\mathrm{U}). A distractor like choice C might misread the atomic number, thinking it's thorium, but subtracting 2 from 94 gives 92 (uranium), not 90—double-check element symbols using periodic table. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 88226Ra→86222Rn+X^{226}_{88}\mathrm{Ra} \rightarrow ^{222}_{86}\mathrm{Rn} + \mathrm{X}. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\mathrm{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (24He^{4}_{2}\mathrm{He}). If atomic number increases by 1 with no mass change → beta emitted (−10e^{0}_{-1}\mathrm{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (01n^{1}_{0}\mathrm{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. Wonderful progress on products!

Question 19

In alpha decay, an alpha particle is emitted. If 88226Ra→ 86222Rn+α^{226}_{88}\mathrm{Ra} \rightarrow \,^{222}_{86}\mathrm{Rn} + \alpha, what is the nuclear notation for the alpha particle α\alpha?

  1. 00γ^{0}_{0}\gamma
  2. 01n^{1}_{0}\mathrm{n}
  3. 24He^{4}_{2}\mathrm{He} (correct answer)
  4. −10e^{0}_{-1}\mathrm{e}
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 238U^{238}\mathrm{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 238U→234Th+4He^{238}\mathrm{U} \rightarrow ^{234}\mathrm{Th} + ^{4}\mathrm{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle ( 4He^{4}\mathrm{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha ( 4He^{4}\mathrm{He}), beta ( −10e^{0}_{-1}\mathrm{e}, electron), neutron ( 1n^{1}\mathrm{n}), proton ( 1H^{1}\mathrm{H}). The equation 88226Ra→86222Rn+α^{226}_{88}\mathrm{Ra} \rightarrow ^{222}_{86}\mathrm{Rn} + \alpha explicitly states alpha decay, and α is the notation for an alpha particle, which is 24He^{4}_{2}\mathrm{He}, a helium nucleus. Choice C correctly interprets the nuclear equation by providing the proper notation for the alpha particle as 24He^{4}_{2}\mathrm{He}. A distractor like choice A might think of gamma as a particle, but gamma is energy ( 00γ^{0}_{0}\gamma) with no mass or charge, not emitted in alpha decay—alpha specifically ejects 24He^{4}_{2}\mathrm{He}. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 226Ra→222Rn+X^{226}\mathrm{Ra} \rightarrow ^{222}\mathrm{Rn} + \mathrm{X}. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\mathrm{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted ( 4He^{4}\mathrm{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\mathrm{e}). If just gamma (energy), no mass or atomic change. If neutrons involved ( 1n^{1}\mathrm{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. Great job recalling that notation!

Question 20

A fission reaction is shown: 92235U+ 01n→ 56144Ba+ 3689Kr+3 01n.^{235}_{92}\mathrm{U} + \,^{1}_{0}\mathrm{n} \rightarrow \,^{144}_{56}\mathrm{Ba} + \,^{89}_{36}\mathrm{Kr} + 3\,^{1}_{0}\mathrm{n}. Which statement best describes the process shown?

  1. Two light nuclei combine to form a heavier nucleus (fusion)
  2. A heavy nucleus splits into smaller nuclei and neutrons (fission) (correct answer)
  3. A nucleus emits an alpha particle (alpha decay)
  4. A nucleus emits only energy with no particle (gamma emission only)
Explanation: This question tests your ability to read and interpret nuclear equations that show how nuclei transform during fission, fusion, or radioactive decay, including identifying particles emitted and products formed. Nuclear equations use special notation where each nucleus or particle is written with its element symbol, mass number (superscript, total protons + neutrons), and sometimes atomic number (subscript, number of protons): for example, 238U^{238}\mathrm{U} represents uranium-238 with mass number 238 and atomic number 92 (uranium always has 92 protons). The equation shows what happens: 238U→234Th+4He^{238}\mathrm{U} \rightarrow ^{234}\mathrm{Th} + ^{4}\mathrm{He} means uranium-238 decays to thorium-234 (mass 234, atomic 90) plus an alpha particle (4He^{4}\mathrm{He}, which is a helium nucleus with mass 4, atomic 2). In nuclear equations, both mass numbers and atomic numbers are conserved (the sums on left equal sums on right): check 238 = 234 + 4 ✓ and 92 = 90 + 2 ✓. Common particles: alpha (4He^{4}\mathrm{He}), beta (−10e^{0}_{-1}\mathrm{e}, electron), neutron (1n^{1}\mathrm{n}), proton (1H^{1}\mathrm{H}). This equation, 92235U+01n→56144Ba+3689Kr+301n^{235}_{92}\mathrm{U} + ^{1}_{0}\mathrm{n} \rightarrow ^{144}_{56}\mathrm{Ba} + ^{89}_{36}\mathrm{Kr} + 3^{1}_{0}\mathrm{n}, shows a heavy uranium nucleus absorbing a neutron and splitting into two smaller nuclei (barium and krypton) plus additional neutrons, with mass (235+1=236, 144+89+3=236) and atomic (92+0=92, 56+36+0=92) conserved. Choice B correctly interprets the nuclear equation by recognizing this as fission, where a heavy nucleus splits. A distractor like choice A might confuse it with fusion, but fusion combines light nuclei, not splits heavy ones—look for splitting and extra neutrons as fission hallmarks. Reading nuclear equations step-by-step: (1) Identify what's on the left (reactant nucleus/nuclei) and right (product nucleus/nuclei and particles). (2) Check conservation: add mass numbers on left, add on right, should equal. Add atomic numbers on left, add on right, should equal. This tells you the equation is valid. (3) Identify unknowns: if you see X in equation, use conservation to find it. Example: 226Ra→222Rn+X^{226}\mathrm{Ra} \rightarrow ^{222}\mathrm{Rn} + X. Mass: 226 = 222 + ? → X has mass 4. Atomic: 88 = 86 + ? → X has atomic 2. So X is 24He^{4}_{2}\mathrm{He} (alpha particle). (4) Classify process: one splitting (fission), two combining (fusion), or one emitting particle (decay). The pattern reveals process type! Common particle recognition: if mass drops by 4 and atomic by 2 → alpha emitted (4He^{4}\mathrm{He}). If atomic number increases by 1 with no mass change → beta emitted (electron, −10e^{0}_{-1}\mathrm{e}). If just gamma (energy), no mass or atomic change. If neutrons involved (1n^{1}\mathrm{n}) and large nucleus splits → fission. If light nuclei combine → fusion. These patterns repeat across nuclear chemistry, so recognizing them once helps with all nuclear equations! Nuclear vs chemical equations: CHEMICAL equations balance atom counts using coefficients (2H2 + O2 → 2H2O keeps atoms same type, just rearranged). NUCLEAR equations balance mass and atomic numbers, but atoms actually CHANGE into different elements (uranium becomes thorium, carbon becomes nitrogen)—this is the key difference! Nuclear changes transform elements, chemical changes only rearrange them. You're building strong skills here!