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AP Physics 2 : AP Physics 2

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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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Example Question #371 : Ap Physics 2

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 activity at 25 minutes after the initial reading.

Possible Answers:

None of these

A=677Bq

A=899Bq

A=150Bq

A=650Bq

Correct answer:

A=650Bq

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.

Rearranging the equation to solve for $\lambda$ .

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

Convert minutes to seconds and plug in values.

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

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

Again use the relationship:

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

Using the new , which is equal to 25min=1500s

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

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 . 15 minutes later, it has an activity of 925 Bq .

Determine the half-life of this isotope.

Possible Answers:

None of these

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

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

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

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

Correct answer:

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

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 pluge in values.

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

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

Use the relationship:

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

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

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

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

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Example Question #12 : Radioactive Nuclear Decay

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 nuclear decay constant of this isotope.

Possible Answers:

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

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

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

None of these

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

Correct answer:

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

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.

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

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

Report an Error

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 . 10 minutes later, it has an activity of 220 Bq .

Determine the half life of this nuclei.

Possible Answers:

282seconds

181seconds

181minutes

None of these

113seconds

Correct answer:

181seconds

Explanation:

Using the relationship:

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

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

Rearranging the equation to solve for $\lambda$ .

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

Converting minutes to seconds and plugging in values.

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

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

Using the relationship

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

Plugging in the calculated value for $\lambda$

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

$181s=t_{1/2}$

Report an Error

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 3.3*10^4Bq . 17 years later, it has an activity of 2.9*10^4 Bq .

Determine the decay constant.

Possible Answers:

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

None of these

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

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

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

Correct answer:

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

Explanation:

Using the relationship:

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

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

Rearranging the equation to solve for $\lambda$ .

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

Converting minutes to seconds and plugging in values.

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

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

Report an Error

Example Question #42 : Quantum And Nuclear Physics

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

Determine the half life.

Possible Answers:

$31.1years=t_{1/2}$

$17.2years=t_{1/2}$

$91.2years=t_{1/2}$

$22.2years=t_{1/2}$

None of these

Correct answer:

$91.2years=t_{1/2}$

Explanation:

Using the relationship:

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

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

Rearranging the equation to solve for $\lambda$ .

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

Converting minutes to seconds and plugging in values.

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

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

Using the relationship

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

Plugging in the calculated value for $\lambda$

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

$91.2years=t_{1/2}$

Report an Error

Example Question #22 : 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 3.3*10^4Bq . 17 years later, it has an activity of 2.9*10^4 Bq .

Determine the number of radioactive atoms in the initial sample.

Possible Answers:

$N=5.71*10^{12}$

$N=3.33*10^{12}$

None of these

$N=5.71*10^{6}$

$N=2.75*10^{12}$

Correct answer:

$N=5.71*10^{12}$

Explanation:

Using the relationship:

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

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

Rearranging the equation to solve for $\lambda$ .

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

Converting minutes to seconds and plugging in values.

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

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

It is then necessary to use the relationship

$A= \lambda N$

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

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

Using the initial activity and the calculated decay constant:

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

$N=5.71*10^{12}$

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Example Question #18 : Radioactive Nuclear Decay

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

Determine the activity 23 years after the original reading.

Possible Answers:

A=3.88*10^4Bq

A=2.77*10^4Bq

A=1.22*10^4Bq

None of these

There will be no activity left

Correct answer:

A=2.77*10^4Bq

Explanation:

Using the relationship:

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

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

Rearranging the equation to solve for $\lambda$ .

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

Converting minutes to seconds and plugging in values.

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

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

Again using the relationship

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

Using the new

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

A=2.77*10^4Bq

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All AP Physics 2 Resources

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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #5 : Radioactive Nuclear Decay

Example Question #31 : Quantum And Nuclear Physics

Example Question #371 : Ap Physics 2

Example Question #371 : Ap Physics 2

Example Question #12 : 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

All AP Physics 2 Resources

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