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AP Physics 2 : Radioactive Nuclear Decay

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Example Questions

Example Question #371 : Ap Physics 2

A scientist takes a reading of a radioactive material, which has an activity of 1570 Bq $\displaystyle 1570 Bq$. 15 minutes later, it has an activity of 925 Bq $\displaystyle 925 Bq$.

Determine the activity at 25 minutes after the initial reading.

Possible Answers:

A=650Bq $\displaystyle A=650Bq$

A=899Bq $\displaystyle A=899Bq$

None of these

A=677Bq $\displaystyle A=677Bq$

A=150Bq $\displaystyle A=150Bq$

Correct answer:

A=650Bq $\displaystyle A=650Bq$

Explanation:

Use the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Here, $\displaystyle A$ is the activity at a given time, A_0 $\displaystyle A_0$ is the initial activity, $\lambda$ $\displaystyle \lambda$ is the radioactive decay constant, and $\displaystyle t$ is the time passed since the initial reading.

Rearranging the equation to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$ $\displaystyle \frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$

$\frac{5.88*10^{-4}}{s}=\lambda$ $\displaystyle \frac{5.88*10^{-4}}{s}=\lambda$

Again use the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Using the new $\displaystyle t$, which is equal to $\displaystyle 25min=1500s$

$A=1570e^{-5.88*10^{-4} 1500}$ $\displaystyle A=1570e^{-5.88*10^{-4} 1500}$

A=650Bq $\displaystyle A=650Bq$

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Example Question #371 : Ap Physics 2

A scientist takes a reading of a radioactive material, which has an activity of 1570 Bq $\displaystyle 1570 Bq$. 15 minutes later, it has an activity of 925 Bq $\displaystyle 925 Bq$.

Determine the half-life of this isotope.

Possible Answers:

None of these

$395s=t_{\frac{1}{2}}$ $\displaystyle 395s=t_{\frac{1}{2}}$

$1007s=t_{\frac{1}{2}}$ $\displaystyle 1007s=t_{\frac{1}{2}}$

$1395s=t_{\frac{1}{2}}$ $\displaystyle 1395s=t_{\frac{1}{2}}$

$1180s=t_{\frac{1}{2}}$ $\displaystyle 1180s=t_{\frac{1}{2}}$

Correct answer:

$1180s=t_{\frac{1}{2}}$ $\displaystyle 1180s=t_{\frac{1}{2}}$

Explanation:

Use the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Rearrange the equation to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and pluge in values.

$\frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$ $\displaystyle \frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$

$\frac{5.88*10^{-4}}{s}=\lambda$ $\displaystyle \frac{5.88*10^{-4}}{s}=\lambda$

Use the relationship:

$\frac{ln(2))}{\lambda}=t_{1/2}$ $\displaystyle \frac{ln(2))}{\lambda}=t_{1/2}$

Plug in the calculated value for $\lambda$ $\displaystyle \lambda$ and solve

$\frac{ln(2))}{5.88*10^{-4}}=t_{\frac{1}{2}}$ $\displaystyle \frac{ln(2))}{5.88*10^{-4}}=t_{\frac{1}{2}}$

$1180s=t_{\frac{1}{2}}$ $\displaystyle 1180s=t_{\frac{1}{2}}$

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Example Question #11 : Radioactive Nuclear Decay

A scientist takes a reading of a radioactive material, which has an activity of 1570 Bq $\displaystyle 1570 Bq$. 15 minutes later, it has an activity of 925 Bq $\displaystyle 925 Bq$.

Determine the nuclear decay constant of this isotope.

Possible Answers:

$\frac{8.55*10^{-4}}{s}=\lambda$ $\displaystyle \frac{8.55*10^{-4}}{s}=\lambda$

$\frac{5.88*10^{-4}}{s}=\lambda$ $\displaystyle \frac{5.88*10^{-4}}{s}=\lambda$

None of these

$\frac{1.13*10^{-4}}{s}=\lambda$ $\displaystyle \frac{1.13*10^{-4}}{s}=\lambda$

$\frac{3.98*10^{-4}}{s}=\lambda$ $\displaystyle \frac{3.98*10^{-4}}{s}=\lambda$

Correct answer:

$\frac{5.88*10^{-4}}{s}=\lambda$ $\displaystyle \frac{5.88*10^{-4}}{s}=\lambda$

Explanation:

Use the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Rearrange the equation to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$ $\displaystyle \frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$

$\frac{5.88*10^{-4}}{s}=\lambda$ $\displaystyle \frac{5.88*10^{-4}}{s}=\lambda$

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Example Question #14 : Radioactive Nuclear Decay

A test is done on a sample of a newly discovered radioactive nuclei, which has an activity of 1990 Bq $\displaystyle 1990 Bq$. $\displaystyle 10 minutes$ later, it has an activity of 220 Bq $\displaystyle 220 Bq$.

Determine the half life of this nuclei.

Possible Answers:

$\displaystyle 282seconds$

$\displaystyle 181seconds$

$\displaystyle 181minutes$

None of these

$\displaystyle 113seconds$

Correct answer:

$\displaystyle 181seconds$

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Here, $\displaystyle A$ is the activity at a given time, A_0 $\displaystyle A_0$ is the intial activity, $\lambda$ $\displaystyle \lambda$ is the radioactive decay constant, and $\displaystyle t$ is the time passed since the initial reading.

Rearranging the equation to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{200}{1990}))}{600s}=\lambda$ $\displaystyle \frac{(-ln(\frac{200}{1990}))}{600s}=\lambda$

$\frac{3.83*10^{-3}}{s}=\lambda$ $\displaystyle \frac{3.83*10^{-3}}{s}=\lambda$

Using the relationship

$\frac{ln(2))}{\lambda}=t_{1/2}$ $\displaystyle \frac{ln(2))}{\lambda}=t_{1/2}$

Plugging in the calculated value for $\lambda$ $\displaystyle \lambda$

$\frac{ln(2))}{3.83*10^{-3}}=t_{1/2}$ $\displaystyle \frac{ln(2))}{3.83*10^{-3}}=t_{1/2}$

$181s=t_{1/2}$ $\displaystyle 181s=t_{1/2}$

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Example Question #21 : Atomic And Nuclear Physics

A scientist discovers a new radioactive nuclei. She runs a test on a sample and finds it has an activity of $\displaystyle 3.3*10^4Bq$. $\displaystyle 17 years$ later, it has an activity of $\displaystyle 2.9*10^4 Bq$.

Determine the decay constant.

Possible Answers:

None of these

$\frac{9.9*10^{-2}}{year}=\lambda$ $\displaystyle \frac{9.9*10^{-2}}{year}=\lambda$

$\frac{7.6*10^{-3}}{year}=\lambda$ $\displaystyle \frac{7.6*10^{-3}}{year}=\lambda$

$\frac{2.2*10^{-5}}{year}=\lambda$ $\displaystyle \frac{2.2*10^{-5}}{year}=\lambda$

$\frac{3.3*10^{-1}}{year}=\lambda$ $\displaystyle \frac{3.3*10^{-1}}{year}=\lambda$

Correct answer:

$\frac{7.6*10^{-3}}{year}=\lambda$ $\displaystyle \frac{7.6*10^{-3}}{year}=\lambda$

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Rearranging the equation to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$ $\displaystyle \frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$

$\frac{7.6*10^{-3}}{year}=\lambda$ $\displaystyle \frac{7.6*10^{-3}}{year}=\lambda$

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Example Question #42 : Quantum And Nuclear Physics

Determine the half life.

Possible Answers:

$31.1years=t_{1/2}$ $\displaystyle 31.1years=t_{1/2}$

$17.2years=t_{1/2}$ $\displaystyle 17.2years=t_{1/2}$

$91.2years=t_{1/2}$ $\displaystyle 91.2years=t_{1/2}$

$22.2years=t_{1/2}$ $\displaystyle 22.2years=t_{1/2}$

None of these

Correct answer:

$91.2years=t_{1/2}$ $\displaystyle 91.2years=t_{1/2}$

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Rearranging the equation to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$ $\displaystyle \frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$

$\frac{7.6*10^{-3}}{year}=\lambda$ $\displaystyle \frac{7.6*10^{-3}}{year}=\lambda$

Using the relationship

$\frac{ln(2))}{\lambda}=t_{1/2}$ $\displaystyle \frac{ln(2))}{\lambda}=t_{1/2}$

Plugging in the calculated value for $\lambda$ $\displaystyle \lambda$

$\frac{ln(2))}{7.6*10^{-3}}=t_{1/2}$ $\displaystyle \frac{ln(2))}{7.6*10^{-3}}=t_{1/2}$

$91.2years=t_{1/2}$ $\displaystyle 91.2years=t_{1/2}$

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Example Question #22 : Atomic And Nuclear Physics

Determine the number of radioactive atoms in the initial sample.

Possible Answers:

$N=5.71*10^{12}$ $\displaystyle N=5.71*10^{12}$

$N=3.33*10^{12}$ $\displaystyle N=3.33*10^{12}$

None of these

$N=5.71*10^{6}$ $\displaystyle N=5.71*10^{6}$

$N=2.75*10^{12}$ $\displaystyle N=2.75*10^{12}$

Correct answer:

$N=5.71*10^{12}$ $\displaystyle N=5.71*10^{12}$

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Rearranging the equation to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$ $\displaystyle \frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$

$\frac{7.6*10^{-3}}{year}=\lambda$ $\displaystyle \frac{7.6*10^{-3}}{year}=\lambda$

It is then necessary to use the relationship

$A= \lambda N$ $\displaystyle A= \lambda N$

Changing the units of the decay constant to be consistent with the activity given.

$\frac{7.6*10^{-3}}{year}*\frac{1year}{365.25days}*\frac{1day}{3600seconds}=\frac{5.78*10^{-9}}{second}$ $\displaystyle \frac{7.6*10^{-3}}{year}*\frac{1year}{365.25days}*\frac{1day}{3600seconds}=\frac{5.78*10^{-9}}{second}$

Using the initial activity and the calculated decay constant:

$3.3*10^{4}= \frac{5.78*10^{-9}}{s} N$ $\displaystyle 3.3*10^{4}= \frac{5.78*10^{-9}}{s} N$

$N=5.71*10^{12}$ $\displaystyle N=5.71*10^{12}$

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Example Question #18 : Radioactive Nuclear Decay

Determine the activity $\displaystyle 23 years$ after the original reading.

Possible Answers:

$\displaystyle A=3.88*10^4Bq$

$\displaystyle A=2.77*10^4Bq$

$\displaystyle A=1.22*10^4Bq$

None of these

There will be no activity left

Correct answer:

$\displaystyle A=2.77*10^4Bq$

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Rearranging the equation to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$ $\displaystyle \frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$

$\frac{7.6*10^{-3}}{year}=\lambda$ $\displaystyle \frac{7.6*10^{-3}}{year}=\lambda$

Again using the relationship

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Using the new $\displaystyle t$

$A=3.3*10^{4}e^{-{7.6*10^{-3} 23}}$ $\displaystyle A=3.3*10^{4}e^{-{7.6*10^{-3} 23}}$

$\displaystyle A=2.77*10^4Bq$

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Example Question #19 : Radioactive Nuclear Decay

Determine the activity $\displaystyle 12 minutes$ after the initial reading.

Possible Answers:

330bq $\displaystyle 330bq$

None of these

83.1bq $\displaystyle 83.1bq$

31.8bq $\displaystyle 31.8bq$

75bq $\displaystyle 75bq$

Correct answer:

31.8bq $\displaystyle 31.8bq$

Explanation:

Using the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Rearranging the equation to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Converting minutes to seconds and plugging in values.

$\frac{(-ln(\frac{200}{1990}))}{600s}=\lambda$ $\displaystyle \frac{(-ln(\frac{200}{1990}))}{600s}=\lambda$

$\frac{3.83*10^{-3}}{s}=\lambda$ $\displaystyle \frac{3.83*10^{-3}}{s}=\lambda$

Again using the relationship

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Using the new $\displaystyle t$, which is equal to $\displaystyle 12min=720seconds$

$A=1990e^{-3.83*10^{-3} 1080}$ $\displaystyle A=1990e^{-3.83*10^{-3} 1080}$

$\displaystyle A=31.8bq$

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Example Question #31 : Atomic And Nuclear Physics

A scientist takes a sample of a newly discovered radioactive element, which has an activity of 1990 Bq $\displaystyle 1990 Bq$. 10 minutes later, it has an activity of 220 Bq $\displaystyle 220 Bq$.

Determine the number of radioactive nuclei in the original sample.

Possible Answers:

None of these

$\displaystyle N=3.77*10^5$

$\displaystyle N=5.42*10^5$

$\displaystyle N=9.85*10^5$

$\displaystyle N=1.44*10^5$

Correct answer:

$\displaystyle N=5.42*10^5$

Explanation:

Use the relationship:

$A=A_0e^{-\lambda t}$ $\displaystyle A=A_0e^{-\lambda t}$

Where $\displaystyle A$ is the activity at a given time, A_0 $\displaystyle A_0$ is the initial activity, $\lambda$ $\displaystyle \lambda$ is the radioactive decay constant and $\displaystyle t$ is the time passed since the initial reading.

Rearrange to solve for $\lambda$ $\displaystyle \lambda$.

$\frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$ $\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\frac{(-ln(\frac{220 Bq}{1990 Bq}))}{600s}=\lambda$ $\displaystyle \frac{(-ln(\frac{220 Bq}{1990 Bq}))}{600s}=\lambda$

$\frac{3.67*10^{-3}}{s}=\lambda$ $\displaystyle \frac{3.67*10^{-3}}{s}=\lambda$

It is then necessary to use the relationship:

$A= \lambda N$ $\displaystyle A= \lambda N$

Use the initial activity and the calculated decay constant:

$1990= \frac{3.67*10^{-3}}{s} N$ $\displaystyle 1990= \frac{3.67*10^{-3}}{s} N$

$\displaystyle N=5.42*10^5$

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AP Physics 2 : Radioactive Nuclear Decay

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #371 : Ap Physics 2

Example Question #371 : Ap Physics 2

Example Question #11 : Radioactive Nuclear Decay

Example Question #14 : Radioactive Nuclear Decay

Example Question #21 : Atomic And Nuclear Physics

Example Question #42 : Quantum And Nuclear Physics

Example Question #22 : Atomic And Nuclear Physics

Example Question #18 : Radioactive Nuclear Decay

Example Question #19 : Radioactive Nuclear Decay

Example Question #31 : Atomic And Nuclear Physics

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