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AP Physics 2 : Radioactive Nuclear Decay

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Example Questions

Example Question #11 : Atomic And Nuclear Physics

Compound X is found to radioactively decay with a rate constant equal to $3.7\cdot 10^{-4}s^{-1}$ . If a time of 20min passes by, what percentage of Compound X has decayed within this period?

Possible Answers:

$36\%$

$50\%$

$75\%$

$25\%$

$64\%$

Correct answer:

$36\%$

Explanation:

To begin with, we'll have to make use of the radioactive decay equation:

$A_{t}=A_{o}e^{-kt}$

Where:

$A_{t}$ = the amount of Compound X after an amount of time, , has passed

$A_{o}$ = the amount of Compound X at the start

= rate constant for the decay process

= the amount of time that has passed

Rearranging, we can see that:

$\frac{A_{t}}{A_{o}}=e^{-kt}$

Furthermore, we can convert minutes into seconds.

$20min\left ( \frac{60s}{1min} \right )=1200s$

Next, we can plug the values we have into the above equation to obtain:

$\frac{A_{t}}{A_{o}}=e^{-kt}=e^{-\left ( 3.7\cdot 10^{-4}s^{-1}\right )(1200s)}=0.64$

From the above expression, we see that:

$\frac{A_{t}}{A_{o}}=0.64$

This means that after a time of 20min has elapsed, there will be $64\%$ as much of Compound X as there was when at the beginning. Since there is $64\%$ remaining after this time period, then we can conclude that $100\%-64\%=36\%$ of Compound X has degraded within this amount of time.

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Example Question #12 : Atomic And Nuclear Physics

Which of the following types of decay particles is known as a positron?

Possible Answers:

$\beta^+$ particle

None of these particles are known as positrons

$\alpha$ particle

$\gamma$ ray

$\beta^-$ particle

Correct answer:

$\beta^+$ particle

Explanation:

Positron decay (positive beta decay) is where a proton is converted into a neutron, a positron, and an electron neutrino.

$p\rightarrow n+e^++v_e$

Positrons are called $\beta^+$ particles (beta plus particles). They are the antiparticle to $\beta^-$ particles (the antimatter version of electrons). They are used in the medical field for PET (Positron Emission Tomography) scans by highlighting a radioactive substance (called a tracer) ingested earlier to show how organs and tissues are working. In nuclear reactors, they cause the water coolant to give a blue glow called Cherenkov radiation. This is because the positrons move faster than light does through water.

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Example Question #1 : Radioactive Nuclear Decay

The half-life of carbon-14 is 5730 years.

Rex the dog died in 1750. What percentage of his original carbon-14 remained in 1975 when he was found by scientists?

Possible Answers:

$77\%$

$97.3\%$

$2.7\%$

$107\%$

$27\%$

Correct answer:

$97.3\%$

Explanation:

225 years have passed since Rex died. Find the number of half-lives that have elapsed.

$\frac{225}{5730}=0.039267$

To find the proportion of a substance that remains after a certain number of half-lives, use the following equation:

$\left(\frac{1}{2} \right )^n$

Here, is the number of half lives that have elapsed.

$\left(\frac{1}{2} \right )^{0.039267}=0.973\cdot 100\%=97.3\%$

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Example Question #3 : Radioactive Nuclear Decay

You measure the beta decay activity of an unknown substance to be 5306Bq . 48 hours later, the activity is 510Bq .

What is the half life in hours?

Possible Answers:

248hr

14.2hr

24hr

1.2hr

16.3hr

Correct answer:

14.2hr

Explanation:

Use the following equation:

$A=A_{0}e^{-\lambda t}$

The decay constant is defined as:

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

From the first equation, we find:

$\lambda=\frac{0.0488}{hr}$

Plug this value into the second equation above and solve.

$t_{\frac{1}{2}}=14.2 hr$

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Example Question #4 : Radioactive Nuclear Decay

A particular sample of a newly discovered isotope has an activity of 1990 Bq . 10 minutes later, it has an activity of 220 Bq .

Determine the radioactive decay constant of this isotope.

Possible Answers:

$\frac{7.51*10^{-2}}{s}=\lambda$

$\frac{7.21*10^{2}}{s}=\lambda$

None of these

$\frac{7.51*10^{-2}}{hr}=\lambda$

$\frac{6.58*10^{-2}}{s}=\lambda$

Correct answer:

$\frac{7.51*10^{-2}}{s}=\lambda$

Explanation:

Use the relationship:

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

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

Rearrange the equation to solve for $\lambda$ .

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

Convert minutes to seconds and plug in values to solve for the radioactive decay constant.

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

$\frac{7.51*10^{-2}}{s}=\lambda$

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Example Question #5 : Radioactive Nuclear Decay

A scientist tests a radioactive sample which has an activity of 1868 Bq . 15 minutes later, it has an activity of 768 Bq .

Determine the number of radioactive nuclei in the initial sample.

Possible Answers:

N=3.58*10^6

N=1.15*10^6

None of these

N=2.15*10^6

N=1.90*10^6

Correct answer:

N=1.90*10^6

Explanation:

Use the relationship:

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

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

Rearrange to solve for $\lambda$ .

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

Convert minutes to seconds and plug in values to find the decay constant.

$\frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\frac{9.86*10^{-4}}{s}=\lambda$

It is then necessary to use the relationship:

$A= \lambda N$

Where is the activity, $\lambda$ is the decay constant and is the number of atoms.

Use the initial activity and the calculated decay constant to solve for the number of atoms:

$1868= \frac{9.84*10^{-4}}{s} N$

N=1.90*10^6

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Example Question #2 : Radioactive Nuclear Decay

A scientist tests a radioactive sample which has an activity of 1868 Bq . 15 minutes later, it has an activity of 768 Bq .

Determine the activity 18 minutes after the initial reading.

Possible Answers:

None of these

A=844Bq

A=714Bq

A=644Bq

A=611Bq

Correct answer:

A=644Bq

Explanation:

Use the relationship:

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

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

Rearrange to solve for $\lambda$

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

Convert minutes to seconds and plug in values.

$\frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\frac{9.86*10^{-4}}{s}=\lambda$

Again use the relationship:

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

Use the new , which is equal to 18min=1080s to plug in and solve for the activity.

$A=1868e^{-9.86*10^{-4} 1080}$

A=644Bq

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Example Question #1 : Radioactive Nuclear Decay

A scientist test a radioactive sample which has an activity of 1868 Bq . 15 minutes later, it has an activity of 768 Bq .

Determine the half life of this isotope.

Possible Answers:

$1138s=t_{\frac{1}{2}}$

None of these

$898s=t_{\frac{1}{2}}$

$702s=t_{\frac{1}{2}}$

$355s=t_{\frac{1}{2}}$

Correct answer:

$702s=t_{\frac{1}{2}}$

Explanation:

Use the relationship:

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

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

Rearrange the equation to solve for $\lambda$ .

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

Convert minutes to seconds and plug in values.

$\frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\frac{9.86*10^{-4}}{s}=\lambda$

Use the relationship:

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

Plug in the calculated value for $\lambda$ :

$\frac{ln(2))}{9.86*10^{-4}}=t_{\frac{1}{2}}$

$702s=t_{\frac{1}{2}}$

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Example Question #5 : Radioactive Nuclear Decay

A scientist test a radioactive sample which has an activity of 1868 Bq . 15 minutes later, it has an activity of 768 Bq .

Determine the nuclear decay constant of this isotope.

Possible Answers:

$\frac{5.55*10^{-4}}{s}=\lambda$

$\frac{9.86*10^{-4}}{s}=\lambda$

None of these

$\frac{4.37*10^{-4}}{s}=\lambda$

$\frac{3.96*10^{-4}}{s}=\lambda$

Correct answer:

$\frac{9.86*10^{-4}}{s}=\lambda$

Explanation:

Use the relationship:

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

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

Rearrange the equation to solve for $\lambda$ .

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

Convert minutes to seconds and plug in values.

$\frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\frac{9.86*10^{-4}}{s}=\lambda$

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Example Question #31 : Quantum And Nuclear Physics

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

Determine the number of radioactive atoms in the initial sample.

Possible Answers:

None of these

N=3.18*10^6

N=6.77*10^6

N=2.67*10^6

N=8.99*10^6

Correct answer:

N=2.67*10^6

Explanation:

Use the relationship:

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

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

Rearrange to solve for $\lambda$ .

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

Convert minutes to seconds and plug in values.

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

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

Use the relationship:

$A= \lambda N$

Where is the activity, $\lambda$ is the decay constant and is the number of nuclei.

Use the initial activity and the calculated decay constant to solve for the number of nuclei:

$1570= \frac{5.88*10^{-4}}{s} N$

N=2.67*10^6

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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 #11 : Atomic And Nuclear Physics

Example Question #12 : Atomic And Nuclear Physics

Example Question #1 : Radioactive Nuclear Decay

Example Question #3 : Radioactive Nuclear Decay

Example Question #4 : Radioactive Nuclear Decay

Example Question #5 : Radioactive Nuclear Decay

Example Question #2 : Radioactive Nuclear Decay

Example Question #1 : Radioactive Nuclear Decay

Example Question #5 : Radioactive Nuclear Decay

Example Question #31 : Quantum And Nuclear Physics

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