IB Physics Quiz: Apply Radioactive Decay

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Apply Radioactive DecayQuestion 1 of 20

A pure sample of Cobalt-60 has an activity of $3.70 \times 10^7$ Bq. The half-life of Cobalt-60 is 5.27 years. What is the approximate mass of the Cobalt-60 in the sample? (1 year ≈ $3.16 \times 10^7$ s, Molar mass of Co-60 ≈ 60 g mol⁻¹, Avogadro's constant ≈ $6.02 \times 10^{23}$ mol⁻¹)

0.89 ng

0.89 µg

0.89 mg

0.89 g

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IB Physics Quiz

IB Physics Quiz: Apply Radioactive Decay

Practice Apply Radioactive Decay in IB Physics 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 Apply Radioactive Decay, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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

0.89 ng
0.89 µg (correct answer)
0.89 mg
0.89 g

Explanation: First, calculate the decay constant

\lambda

in s⁻¹.

T_{1/2} = 5.27 \text{ yr} \times 3.16 \times 10^7 \text{ s/yr} \approx 1.665 \times 10^8 \text{ s}

\lambda = \ln(2) / T_{1/2} \approx 0.693 / (1.665 \times 10^8 \text{ s}) \approx 4.16 \times 10^{-9} \text{ s}^{-1}

. Next, find the number of nuclei

N

using

A = \lambda N

N = A / \lambda = (3.70 \times 10^7 \text{ Bq}) / (4.16 \times 10^{-9} \text{ s}^{-1}) \approx 8.89 \times 10^{15}

nuclei. Finally, convert nuclei to mass. Mass =

(N / N_A) \times \text{Molar Mass} = (8.89 \times 10^{15} / 6.02 \times 10^{23} \text{ mol}^{-1}) \times 0.060 \text{ kg mol}^{-1} \approx 8.86 \times 10^{-10} \text{ kg} = 0.886 \text{ µg}

Question 2

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?

The nucleus will decay at exactly time $T_{1/2}$ .
The nucleus has a 50% chance of having decayed after one half-life has passed. (correct answer)
The probability of the nucleus decaying per unit time is constant and equal to $1/T_{1/2}$ .
The nucleus is certain to have decayed by time $2T_{1/2}$ .

Explanation: 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.

Question 3

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?

5.41 MeV
5.30 MeV (correct answer)
2.71 MeV
0.11 MeV

Explanation: 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}

Question 4

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?

Alpha radiation, because it is highly ionizing and creates a strong signal in the detector before the foil.
Beta radiation, because its penetration is sufficient to pass through the foil but is sensitive to thickness changes. (correct answer)
Gamma radiation, because it has the highest penetrating power and will not be significantly stopped by the foil.
Any type of radiation would work, as long as the source has a high enough initial activity.

Explanation: 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.

Question 5

A sample is prepared with an equal number of nuclei of isotope X and isotope Y. The half-life of X is $T$ , and the half-life of Y is $2T$ . What is the ratio of the activity of X to the activity of Y, $A_X / A_Y$ , at time $t=4T$ ?

1/4
1/2 (correct answer)
1
2

Explanation: Activity

A = \lambda N

. First find the decay constants:

\lambda_X = \ln(2)/T

and

\lambda_Y = \ln(2)/(2T)

, so

\lambda_X = 2\lambda_Y

. Next, find the number of nuclei remaining at

t=4T

. For X, 4 half-lives pass:

N_X = N_0(1/2)^4 = N_0/16

. For Y, 2 half-lives pass:

N_Y = N_0(1/2)^2 = N_0/4

. Now find the ratio of activities:

A_X/A_Y = (\lambda_X N_X)/(\lambda_Y N_Y) = (2\lambda_Y \times N_0/16)/(\lambda_Y \times N_0/4) = (2/16)/(1/4) = (1/8)/(1/4) = 4/8 = 1/2

Question 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?

The half-life of the isotope.
The initial activity of the sample.
The decay constant of the isotope. (correct answer)
The average lifetime of the isotope.

Explanation: 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.

Question 7

A sample contains two radioactive isotopes, P and Q. Isotope P has a half-life of 10 minutes. Isotope Q has a half-life of 30 minutes. Initially, the activity of P is equal to the activity of Q. Which statement correctly describes the total activity of the sample over the next hour?

The total activity will decrease with a constant effective half-life of 20 minutes.
The total activity will decrease, and its effective half-life will gradually increase. (correct answer)
The total activity will decrease, and its effective half-life will gradually decrease.
The total activity will remain approximately constant as the two decays balance each other.

Explanation: Initially, the total activity is dominated by the decay of the shorter-lived isotope, P. The initial rate of decrease of total activity will be fast, corresponding to an effective half-life closer to 10 minutes. As time passes, isotope P decays away much more quickly than Q. After a significant amount of time (e.g., an hour), the sample will consist mainly of isotope Q, and the total activity will be almost entirely due to Q. Therefore, the rate of decay will slow down, and the effective half-life of the mixture will approach 30 minutes. The effective half-life thus increases over time.

Question 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?

Beta radiation, with a continuous energy spectrum.
Beta radiation, with a discrete energy spectrum.
Gamma radiation, with a continuous energy spectrum.
Gamma radiation, with a discrete energy spectrum. (correct answer)

Explanation: 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.

Question 9

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)

5.8 × 10⁵ Bq (correct answer)
4.6 × 10⁶ Bq
1.5 × 10⁶ Bq
3.7 × 10⁷ Bq

Explanation:

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.

Question 10

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?

19.5 s⁻¹
20.0 s⁻¹
20.5 s⁻¹ (correct answer)
200.5 s⁻¹

Explanation: 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}

Question 11

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?

Beta-minus decay, which increases Z by one. (correct answer)
Beta-minus decay, which decreases Z by one.
Beta-plus decay, which increases Z by one.
Beta-plus decay, which decreases Z by one.

Explanation: 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.

Question 12

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?

6 alpha and 4 beta-minus decays. (correct answer)
6 alpha and 8 beta-minus decays.
8 alpha and 6 beta-minus decays.
24 alpha and 8 beta-minus decays.

Explanation: The change in mass number

\Delta A

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.

Question 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?

The electrostatic repulsion between the two protons is much stronger in Helium-5 than in Helium-4.
The strong nuclear force does not act on the third neutron in Helium-5, leaving it unbound.
The additional neutron in Helium-5 must occupy a higher energy level and does not add sufficient binding energy. (correct answer)
Helium-5 can only be formed in high-energy collisions, whereas Helium-4 is a natural decay product.

Explanation: 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.

Question 14

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?

2850 years
5700 years
11400 years (correct answer)
22800 years

Explanation: 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}

Question 15

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 ( $N_D$ )?

The number of daughter nuclei $N_D$ is approximately equal to the number of parent nuclei $N_P$ .
The activity of the daughter $A_D$ is much less than the activity of the parent $A_P$ .
The number of daughter nuclei $N_D$ is much greater than the number of parent nuclei $N_P$ .
The activity of the daughter $A_D$ is approximately equal to the activity of the parent $A_P$ . (correct answer)

Explanation: 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

Question 16

The binding energy per nucleon of Uranium-238 is approximately 7.6 MeV, while the binding energy per nucleon for its alpha decay product, Thorium-234, is slightly higher. How does this difference in binding energy account for the energy released during the decay?

The alpha particle is created from pure energy, which is subtracted from the total binding energy of the Uranium nucleus.
The binding energy of the Uranium nucleus is released as kinetic energy of the alpha particle, leaving the Thorium nucleus with less energy.
The mass of the Uranium nucleus is converted entirely into the binding energy of the Thorium nucleus.
The total binding energy of the products (Thorium nucleus and alpha particle) is greater than the binding energy of the Uranium nucleus. (correct answer)

Explanation: Energy is released in a nuclear reaction if the products are more stable (more tightly bound) than the reactants. This means the total binding energy of the products is greater than the total binding energy of the original nucleus. The binding energy of the emitted alpha particle is very high (about 28 MeV). The sum of the binding energies of the Th-234 nucleus and the alpha particle is greater than the binding energy of the U-238 nucleus. This increase in total binding energy corresponds to a decrease in total mass (mass defect), and this lost mass is converted into the kinetic energy of the decay products according to

E=mc^2

Question 17

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?

8 days
12 days (correct answer)
16 days
24 days

Explanation: 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.

Question 18

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?

15 minutes (correct answer)
20 minutes
25 minutes
30 minutes

Explanation: 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.

Question 19

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?

12 days
18 days
15 days (correct answer)
24 days

Explanation: 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.

Question 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?

18%
75%
82% (correct answer)
88%

Explanation: 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%.