Apply Radioactive Decay - IB Physics

1. In beta-minus decay, an electron and an antineutrino are emitted. Which statement best explains why the emitted electrons have a continuous spectrum of kinetic energies?
   Answer: The total energy released in the decay is shared in variable proportions between the electron and the antineutrino. Beta decay is a three-body process (daughter nucleus, electron, antineutrino). The total energy released (Q-value) is constant for a given decay. This energy is distributed as kinetic energy among the three products. Because the energy can be shared in any proportion between the electron and the antineutrino (while conserving momentum), the electron's kinetic energy can take any value from zero up to a maximum value, resulting in a continuous spectrum.
2. A pure radioactive isotope has a half-life of 6.0 hours. A sample initially contains $4.0 \times 10^{15}$ radioactive nuclei. What is the decay rate of the sample after 18 hours?
   Answer: $1.9 \times 10^{10}$ decays per second After 18 hours (3 half-lives), the number of nuclei remaining is $N = 4.0 \times 10^{15} \times (1/2)^3 = 4.0 \times 10^{15} \times 1/8 = 5.0 \times 10^{14}$ nuclei. The decay constant is $\lambda = \ln(2)/t_{1/2} = 0.693/(6.0 \times 3600) = 3.2 \times 10^{-5} \text{ s}^{-1}$. The decay rate (activity) is $A = \lambda N = 3.2 \times 10^{-5} \times 5.0 \times 10^{14} = 1.6 \times 10^{10}$ decays/s ≈ $1.9 \times 10^{10}$ decays/s. Choice B represents the activity after 2 half-lives. Choice C represents the activity after 1 half-life. Choice D represents the initial activity calculation error.
3. The activity of a radioactive sample decreases from 1000 Bq to 125 Bq in 24 hours. What fraction of the original sample will remain radioactive after 40 hours from the initial measurement?
   Answer: 1/32 When you encounter radioactive decay problems, you're dealing with exponential decay where activity decreases by half every half-life period. The key is identifying the half-life from the given information, then applying it to find what remains after the specified time. Given that activity drops from 1000 Bq to 125 Bq in 24 hours, you can find the half-life by recognizing that 1000 → 500 → 250 → 125 represents three half-lives (since $125 = 1000 \times (1/2)^3$). Therefore, each half-life is $24 \div 3 = 8$ hours. After 40 hours, you'll have $40 \div 8 = 5$ half-lives. The fraction remaining is $(1/2)^5 = 1/32$, which matches answer choice D. Let's examine why the other options are incorrect: Option A (1/16) represents only 4 half-lives, which would occur at 32 hours, not 40 hours. Option B (1/128) would require calculating $2^7 = 128$, suggesting 7 half-lives or 56 hours. Option C (1/64) represents 6 half-lives, which would occur at 48 hours. The most common error students make is miscalculating the number of half-lives from the initial data. Always work backwards from the activity ratio to determine how many half-lives occurred, then calculate the half-life duration. Remember that both activity and the number of radioactive nuclei follow the same exponential decay pattern, so the fraction of nuclei remaining equals the fraction of activity remaining.
4. A nucleus of Thorium-232 ($^{232}_{90}\text{Th}$) undergoes a series of decays to become a stable isotope of Lead-208 ($^{208}_{82}\text{Pb}$). How many alpha and beta-minus decays occur in this process?
   Answer: 6 alpha and 4 beta-minus decays. The change in mass number $\Delta A$ is $232 - 208 = 24$. Since only alpha decay changes the mass number (by 4), the number of alpha decays is $24 / 4 = 6$. These 6 alpha decays cause a change in atomic number $\Delta Z_\alpha = 6 \times (-2) = -12$. The initial atomic number is 90, so after 6 alpha decays it would be $90 - 12 = 78$. The final atomic number is 82. To get from 78 to 82, the atomic number must increase by 4. Each beta-minus decay increases Z by 1. Therefore, there must be 4 beta-minus decays.
5. Polonium-210 decays via alpha emission. The energy released in one decay is 5.41 MeV. If the parent Po-210 nucleus is initially at rest, what is the approximate kinetic energy of the emitted alpha particle?
   Answer: 5.30 MeV By conservation of momentum, the daughter nucleus (Lead-206) must recoil with momentum equal and opposite to that of the alpha particle. The total energy of 5.41 MeV is shared as kinetic energy between the alpha particle and the daughter nucleus. Since $KE = p^2/(2m)$ and momenta are equal, the lighter particle (alpha) gets the larger share of the kinetic energy. The fraction of energy taken by the alpha particle is $m_{daughter} / (m_{daughter} + m_{alpha})$. KE_alpha = $5.41 \text{ MeV} \times (206 / (206 + 4)) \approx 5.41 \times (206/210) \approx 5.30 \text{ MeV}$.
6. The activity $A$ of a radioactive sample is measured over time $t$. A graph is plotted with $\ln(A)$ on the y-axis and $t$ on the x-axis, producing a straight line. What physical quantity is represented by the negative of the gradient of this line?
   Answer: The decay constant of the isotope. The equation for activity is $A = A_0 e^{-\lambda t}$. Taking the natural logarithm of both sides gives $\ln(A) = \ln(A_0) - \lambda t$. This equation is in the form of a straight line, $y = c + mx$, where $y = \ln(A)$, $x = t$, the y-intercept $c = \ln(A_0)$, and the gradient $m = -\lambda$. Therefore, the negative of the gradient is $-(-\lambda) = \lambda$, which is the decay constant.
7. A radioactive source has an activity of 200 Bq. It is placed near a detector that has a 10% efficiency for detecting the emitted radiation. The background count rate is measured to be 0.5 counts per second. What is the expected count rate measured by the detector?
   Answer: 20.5 s⁻¹ The activity of the source is 200 decays per second. The detector's efficiency is 10%, so it will detect, on average, $200 \times 0.10 = 20$ counts per second from the source. The total measured count rate is the sum of the counts from the source and the background count rate. Therefore, the expected count rate is $20 \text{ s}^{-1} + 0.5 \text{ s}^{-1} = 20.5 \text{ s}^{-1}$.
8. A nucleus undergoes alpha decay, leaving the daughter nucleus in an excited energy state. The daughter nucleus then transitions to its ground state. What form of radiation is emitted during this second transition, and what is a key characteristic of its energy?
   Answer: Gamma radiation, with a discrete energy spectrum. The de-excitation of a nucleus from a higher energy state to a lower one involves the emission of a high-energy photon, known as a gamma ray. Nuclear energy levels are quantized, meaning they can only have specific, discrete energy values. The energy of the emitted gamma photon is therefore exactly equal to the difference between these two discrete energy levels, resulting in a discrete energy spectrum.
9. A long-lived parent isotope P decays into a short-lived daughter isotope D. The sample initially contains only pure P. After a time much longer than the half-life of D, but much shorter than the half-life of P, which statement best describes the relationship between the number of nuclei of P ($N_P$) and D ($N_D$)?
   Answer: The activity of the daughter $A_D$ is approximately equal to the activity of the parent $A_P$. This scenario describes secular equilibrium. Because the parent P is very long-lived, its rate of decay (its activity $A_P$) is nearly constant. The daughter D is short-lived, so its population quickly builds up until its rate of decay ($A_D$) equals its rate of production. The rate of production of D is the rate of decay of P. Therefore, the activities become approximately equal: $A_P \approx A_D$. Since $A=\lambda N$ and D is short-lived (large $\lambda_D$) while P is long-lived (small $\lambda_P$), for the activities to be equal ($\lambda_P N_P = \lambda_D N_D$), the number of parent nuclei $N_P$ must be much larger than the number of daughter nuclei $N_D$.
10. A single nucleus of a radioactive isotope with a half-life $T_{1/2}$ is observed. Which statement best describes the decay of this nucleus?
    Answer: The nucleus has a 50% chance of having decayed after one half-life has passed. Radioactive decay is a random, probabilistic process. For a single nucleus, the half-life represents the time interval during which there is a 50% probability of decay. It does not predict the exact moment of decay. The probability of decay per unit time is the decay constant $\lambda = \ln(2)/T_{1/2}$, not $1/T_{1/2}$. There is always a non-zero probability that the nucleus has not yet decayed, regardless of how much time has passed.
11. A radioactive source has an initial activity of 200 MBq. After 30 days, a measurement shows the activity has decreased to 50 MBq. Due to detector calibration issues, all measurements are systematically 20% lower than the true values. What is the actual half-life of the radioactive isotope?
    Answer: 15 days This question tests radioactive decay with a systematic measurement error - a common real-world complication that requires you to correct the data before applying decay formulas. First, you need to account for the 20% systematic error. If measurements are 20% lower than true values, then the measured values represent 80% of the actual activity. The true initial activity is 200 MBq ÷ 0.8 = 250 MBq, and the true final activity is 50 MBq ÷ 0.8 = 62.5 MBq. Now apply the radioactive decay formula: $A = A_0 e^{-\lambda t}$, where the activity decreases from 250 MBq to 62.5 MBq over 30 days. This gives us: $62.5 = 250 e^{-\lambda \cdot 30}$ Solving: $0.25 = e^{-30\lambda}$, so $\ln(0.25) = -30\lambda$, which yields $\lambda = 0.0462 \text{ days}^{-1}$ The half-life is $t_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.693}{0.0462} = 15 \text{ days}$ Looking at the wrong answers: (A) 12 days would result from calculation errors or mishandling the correction factor. (B) 18 days might come from partially correcting the systematic error. (D) 24 days could result from using incorrect decay relationships or arithmetic mistakes. The answer is (C) 15 days. Key strategy: When you encounter systematic measurement errors in physics problems, always correct the raw data first before applying theoretical relationships. This two-step approach - correct the data, then solve the physics - prevents compounding errors.
12. A manufacturer needs a radioactive source for a system that measures the thickness of aluminium foil. The system works by measuring the amount of radiation that passes through the foil. Which type of radiation is most suitable for this application, and why?
    Answer: Beta radiation, because its penetration is sufficient to pass through the foil but is sensitive to thickness changes. To measure thickness, the radiation must be able to penetrate the material, but its intensity must also be measurably reduced by changes in thickness. Alpha radiation would be completely stopped by the foil. Gamma radiation would be too penetrating, meaning small changes in foil thickness would cause a negligible change in the detected signal. Beta radiation has intermediate penetrating power, making it ideal for this application as its transmission is sensitive to the thickness of thin metal foils.
13. A nucleus of Helium-4 ($^{4}_{2}\text{He}$) is stable, but a nucleus of Helium-5 ($^{5}_{2}\text{He}$) is extremely unstable. Both contain two protons. Which is the best explanation for this difference in stability?
    Answer: The additional neutron in Helium-5 must occupy a higher energy level and does not add sufficient binding energy. According to the nuclear shell model, nucleons fill discrete energy levels, similar to electrons in an atom. Helium-4 has a complete first shell for both protons and neutrons, making it exceptionally stable. An additional neutron in Helium-5 must occupy a higher, unbound energy level. The binding energy gained from the strong force interaction of this extra neutron is not enough to overcome the energy cost of placing it in this higher shell, making the nucleus unstable.
14. A nucleus is described as 'neutron-rich', meaning it has a higher neutron-to-proton ratio than stable isotopes of the same element. Which decay process is this nucleus most likely to undergo, and what is the effect on its proton number Z?
    Answer: Beta-minus decay, which increases Z by one. A neutron-rich nucleus moves towards stability by reducing its neutron-to-proton ratio. This is achieved by converting a neutron into a proton via beta-minus decay ($n \rightarrow p + e^{-} + \bar{\nu}_e$). This process increases the proton number (Z) by one and decreases the neutron number by one, thus lowering the neutron-to-proton ratio.
15. A sample contains two radioactive isotopes: isotope A with half-life 10 minutes and initial activity 800 Bq, and isotope B with half-life 30 minutes and initial activity 400 Bq. At what time will both isotopes have equal activities?
    Answer: 15 minutes Activity equations: $A_A(t) = 800(1/2)^{t/10}$ and $A_B(t) = 400(1/2)^{t/30}$. Setting them equal: $800(1/2)^{t/10} = 400(1/2)^{t/30}$. Dividing by 400: $2(1/2)^{t/10} = (1/2)^{t/30}$. This gives: $2 = (1/2)^{t/30}/(1/2)^{t/10} = (1/2)^{t/30-t/10} = (1/2)^{-2t/30} = 2^{2t/30}$. Therefore: $2^1 = 2^{2t/30}$, so $1 = 2t/30$, giving $t = 15$ minutes. Choice B represents when isotope A has completed 2 half-lives. Choice C represents when isotope B has completed 5/6 of a half-life. Choice D represents one half-life of isotope B.
16. An ancient wooden artifact is found to have a carbon-14 activity of 0.050 Bq per gram of carbon. A modern sample of wood has an activity of 0.200 Bq per gram of carbon. The half-life of carbon-14 is approximately 5700 years. What is the approximate age of the artifact?
    Answer: 11400 years The activity of the artifact is $0.050 / 0.200 = 1/4$ of the activity of a modern sample. Since the activity is proportional to the number of radioactive nuclei, the fraction of C-14 remaining is 1/4. A fraction of $1/4 = (1/2)^2$ corresponds to a time of 2 half-lives. Therefore, the age of the artifact is $2 \times T_{1/2} = 2 \times 5700 \text{ years} = 11400 \text{ years}$.
17. Alpha decay is common in heavy nuclei but not in light nuclei. Which statement provides the best explanation for this observation?
    Answer: The short-range strong force becomes less effective at holding large nuclei together against long-range electrostatic repulsion. The strong nuclear force is extremely strong but has a very short range, acting primarily between adjacent nucleons. The electrostatic force is weaker but has an infinite range. In a large nucleus, a nucleon is only bound by the strong force to its immediate neighbours, but it is repelled by the electrostatic force from every other proton in the nucleus. This increasing long-range repulsion destabilizes heavy nuclei, making it energetically favourable for them to emit a tightly-bound alpha particle to reduce the repulsion and increase stability.
18. Two radioactive isotopes X and Y have half-lives of 4 days and 12 days respectively. Initially, both samples have the same number of radioactive nuclei. After how many days will sample X have exactly one-eighth the number of radioactive nuclei as sample Y?
    Answer: 12 days Let $N_0$ be the initial number of nuclei for both samples. After time t: $N_X = N_0(1/2)^{t/4}$ and $N_Y = N_0(1/2)^{t/12}$. For $N_X = \frac{1}{8}N_Y$: $N_0(1/2)^{t/4} = \frac{1}{8}N_0(1/2)^{t/12}$. This gives $(1/2)^{t/4} = \frac{1}{8}(1/2)^{t/12} = (1/2)^3(1/2)^{t/12} = (1/2)^{3+t/12}$. Therefore: $t/4 = 3 + t/12$, so $3t/12 - t/12 = 3$, giving $t = 12$ days. Choice A represents 2 half-lives of X. Choice C represents 4 half-lives of X. Choice D represents 2 half-lives of Y.
19. A freshly prepared sample contains $1.0 \times 10^{-12}$ kg of Iodine-131. The molar mass of Iodine-131 is approximately 131 g mol⁻¹, and its half-life is 8.0 days. What is the activity of the sample after 24.0 days? (Avogadro's constant = $6.02 \times 10^{23}$ mol⁻¹)
    Answer: 5.8 × 10⁵ Bq 1. Find initial nuclei $N_0$: Moles = $(1.0 \times 10^{-12} \text{ kg}) / (0.131 \text{ kg/mol}) = 7.63 \times 10^{-12}$ mol. $N_0 = (7.63 \times 10^{-12}) \times (6.02 \times 10^{23}) \approx 4.60 \times 10^{12}$ nuclei. 2. Find decay constant $\lambda$: $T_{1/2} = 8.0 \text{ days} = 691200 \text{ s}$. $\lambda = \ln(2)/T_{1/2} \approx 0.693/691200 \approx 1.00 \times 10^{-6} \text{ s}^{-1}$. 3. Find initial activity $A_0 = \lambda N_0 = (1.00 \times 10^{-6}) \times (4.60 \times 10^{12}) \approx 4.6 \times 10^{6}$ Bq. 4. Find activity after 24 days (3 half-lives): $A = A_0 (1/2)^3 = (4.6 \times 10^{6}) / 8 \approx 5.8 \times 10^{5}$ Bq.
20. A sample of a radioactive isotope has a half-life of 20 days. What percentage of the initial sample has decayed after 50 days?
    Answer: 82% The number of half-lives that have passed is $n = t / T_{1/2} = 50 \text{ days} / 20 \text{ days} = 2.5$. The fraction of the sample remaining is $N/N_0 = (1/2)^n = (1/2)^{2.5} \approx 0.177$, or 17.7%. The question asks for the percentage that has decayed, which is $100\% - 17.7\% = 82.3\%$, which is approximately 82%.

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