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AP Physics 2 : Quantum and Nuclear Physics

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

Example Questions

Example Question #381 : Ap Physics 2

A scientist is working with a new radioactive nuclei. A sample has an activity of 199 Bq . 8 minutes later, it has an activity of 9 Bq .

Determine the nuclear decay constant.

Possible Answers:

None of these

$\frac{2.22*10^{-3}}{s}=\lambda$

$\frac{5.55*10^{-3}}{s}=\lambda$

$\frac{3.77*10^{-3}}{s}=\lambda$

$\frac{6.45*10^{-3}}{s}=\lambda$

Correct answer:

$\frac{6.45*10^{-3}}{s}=\lambda$

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.

$\frac{(-ln(\frac{9 Bq}{199 Bq}))}{480s}=\lambda$

$\frac{6.45*10^{-3}}{s}=\lambda$

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Example Question #32 : Atomic And Nuclear Physics

A scientist is working with a new radioactive nuclei. A sample has an activity of 199 Bq . 8 minutes later, it has an activity of 9 Bq .

Determine the activity 4 minutes after the initial reading.

Possible Answers:

A=2Bq

A=25Bq

None of these

A=42Bq

A=88Bq

Correct answer:

A=42Bq

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.

$\frac{(-ln(\frac{9 Bq}{199 Bq}))}{480s}=\lambda$

$\frac{6.45*10^{-3}}{s}=\lambda$

Again, use the relationship:

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

Using the new , which is equal to 4min=240s

$A=199e^{-6.45*10^{-3} 240}$

A=42Bq

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Example Question #33 : Atomic And Nuclear Physics

A scientist is working with a new radioactive nuclei. A sample has an activity of 199 Bq . 8 minutes later, it has an activity of 9 Bq .

Determine the number of radioactive atoms at the time of the initial reading.

Possible Answers:

N=3.01*10^4

N=8.87*10^4

N=2.34*10^4

None of these

N=8.11*10^4

Correct answer:

N=3.01*10^4

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.

$\frac{(-ln(\frac{9 Bq}{199 Bq}))}{480s}=\lambda$

$\frac{6.45*10^{-3}}{s}=\lambda$

It is then necessary to use the relationship:

$A= \lambda N$

Use the initial activity and the calculated decay constant:

$199= \frac{6.45*10^{-3}}{s} N$

N=3.01*10^4

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Example Question #34 : Atomic And Nuclear Physics

A scientist runs a test on a radioactive sample and finds that it has an activity of 6.0*10^3 Bq . Three days later, she runs another test and finds the activity to be 2.9*10^3Bq .

Another sample is tested. It has an activity of 3.8*10^4Bq . Determine the number of radioactive nuclei in it.

Possible Answers:

$N=1.36*10^{10}$

$N=7.75*10^{10}$

$N=1.36*10^{23}$

None of these

$N=1.12*10^{10}$

Correct answer:

$N=1.36*10^{10}$

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 days to seconds and plug in values.

$\frac{(-ln(\frac{2.9*10^3 Bq}{6.0*10^3 Bq}))}{2.6*10^5s}=\lambda$

$\frac{2.8*10^{-6}}{s}=\lambda$

It is then necessary to use the relationship:

$A= \lambda N$

Use the initial activity and the calculated decay constant:

$3.8*10^4= \frac{2.8*10^{-6}}{s} N$

$N=1.36*10^{10}$

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Example Question #35 : Atomic And Nuclear Physics

A scientist runs a test on a radioactive sample and finds that it has an activity of 6.0*10^3 Bq . Three days later, she runs another test and finds the activity to be 2.9*10^3Bq .

Determine the number of radioactive nuclei in the original sample.

Possible Answers:

None of these

N=1.44*10^9

N=2.14*10^9

N=3.88*10^9

N=9.61*10^9

Correct answer:

N=2.14*10^9

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 days to seconds and plug in values.

$\frac{(-ln(\frac{2.9*10^3 Bq}{6.0*10^3 Bq}))}{2.6*10^5s}=\lambda$

$\frac{2.8*10^{-6}}{s}=\lambda$

It is then necessary to use the relationship:

$A= \lambda N$

Use the initial activity and the calculated decay constant:

$6.0*10^3= \frac{2.8*10^{-6}}{s} N$

N=2.14*10^9

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Example Question #36 : Atomic And Nuclear Physics

A scientist runs a test on a radioactive sample and finds that it has an activity of 6.0*10^3 Bq . Three days later, she runs another test and finds the activity to be 2.9*10^3Bq .

Determine the half-life.

Possible Answers:

None of these

$2.48*10^5s=t_{1/2}$

$7.85*10^5s=t_{1/2}$

$8.35*10^5s=t_{1/2}$

$3.34*10^5s=t_{1/2}$

Correct answer:

$2.48*10^5s=t_{1/2}$

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 days to seconds and plug in values.

$\frac{(-ln(\frac{2.9*10^3 Bq}{6.0*10^3 Bq}))}{2.6*10^5s}=\lambda$

$\frac{2.8*10^{-6}}{s}=\lambda$

Use the relationship:

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

Plugin the calculated value for $\lambda$ :

$\frac{ln(2))}{2.8*10^{-6}}=t_{1/2}$

$2.48*10^5s=t_{1/2}$

Report an Error

Example Question #382 : Ap Physics 2

A scientist is working with a new radioactive nuclei. A sample has an activity of 199 Bq . 8 minutes later, it has an activity of 9 Bq .

Determine the half life.

Possible Answers:

$28s=t_{1/2}$

$107s=t_{1/2}$

$1000s=t_{1/2}$

$77s=t_{1/2}$

None of these

Correct answer:

$107s=t_{1/2}$

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.

$\frac{(-ln(\frac{9 Bq}{199 Bq}))}{480s}=\lambda$

$\frac{6.45*10^{-3}}{s}=\lambda$

Use the relationship:

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

Plug in the calculated value for $\lambda$

$\frac{ln(2))}{6.45*10^{-3}}=t_{1/2}$

$107s=t_{1/2}$

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Example Question #38 : Atomic And Nuclear Physics

A scientist runs a test on a radioactive sample and finds that it has an activity of 6.0*10^3 Bq . Three days later, she runs another test and finds the activity to be 2.9*10^3Bq .

Determine the activity of the sample 12 hours after the original reading.

Possible Answers:

A=5316Bq

None of these

A=4014Bq

A=4449Bq

A=6706Bq

Correct answer:

A=5316Bq

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 days to seconds and plug in values.

$\frac{(-ln(\frac{2.9*10^3 Bq}{6.0*10^3 Bq}))}{2.6*10^5s}=\lambda$

$\frac{2.8*10^{-6}}{s}=\lambda$

Again, use the relationship:

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

Use the new , which is equal to 18min=1080s

$A=6.0*10^3e^{-2.8*10^{-6} 4.32*10^4}$

A=5316Bq

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Example Question #39 : Atomic And Nuclear Physics

A particular radioactive sample initially has $20*10^{23}$ atoms and a half-life of 30min. How many atoms are there after 2hr?

Possible Answers:

$1.25 \times 10^{23}atoms$

$10.0 \times 10^{23}atoms$

$2.5 \times 10^{23}atoms$

$5.0 \times 10^{23}atoms$

$25.0 \times 10^{23}atoms$

Correct answer:

$1.25 \times 10^{23}atoms$

Explanation:

After 1 half-life of a radioactive sample has passed, only half of the initial particles will remain. We can see that 2 hours contains four 30-minute half-life periods. Therefore, through using our equation for radioactive decay:

$A_n=A_0\left(\frac{1}{2} \right )^n$

$A_n=20 * 10^{23}atoms\left(\frac{1}{2} \right )^4=1.25* 10^{23}atoms$

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Example Question #392 : Ap Physics 2

A new element is discovered named Banfordium. A test is run on a sample and finds it has an activity of $4.28*10^{13} Bq$ . Three years later, the same sample has an activity of $8.78*10^{12} Bq$ .

Determine the decay constant of Banfordium.

Possible Answers:

$\frac{.711}{year}$

$\frac{.525}{year}$

$\frac{.825}{year}$

None of these

$\frac{.448}{year}$

Correct answer:

$\frac{.525}{year}$

Explanation:

Calculate the decay constant:

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

Solve for $\lambda$

$\lambda=-ln{\frac{A}{A_0}}*\frac{1}{t}$

Plug in values:

$\lambda=-ln{\frac{8.7*10^{12}}{4.2*10^{13}}}*\frac{1}{3}$

$\lambda=\frac{.525}{year}$

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AP Physics 2 : Quantum and Nuclear Physics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #381 : Ap Physics 2

Example Question #32 : Atomic And Nuclear Physics

Example Question #33 : Atomic And Nuclear Physics

Example Question #34 : Atomic And Nuclear Physics

Example Question #35 : Atomic And Nuclear Physics

Example Question #36 : Atomic And Nuclear Physics

Example Question #382 : Ap Physics 2

Example Question #38 : Atomic And Nuclear Physics

Example Question #39 : Atomic And Nuclear Physics

Example Question #392 : Ap Physics 2

All AP Physics 2 Resources

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