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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 #21 : Radioactive Nuclear Decay

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

Possible Answers:

$5.0 \times 10^{23}atoms$ $\displaystyle 5.0 \times 10^{23}atoms$

$1.25 \times 10^{23}atoms$ $\displaystyle 1.25 \times 10^{23}atoms$

$10.0 \times 10^{23}atoms$ $\displaystyle 10.0 \times 10^{23}atoms$

$25.0 \times 10^{23}atoms$ $\displaystyle 25.0 \times 10^{23}atoms$

$2.5 \times 10^{23}atoms$ $\displaystyle 2.5 \times 10^{23}atoms$

Correct answer:

$1.25 \times 10^{23}atoms$ $\displaystyle 1.25 \times 10^{23}atoms$

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:

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

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

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Example Question #391 : Ap Physics 2

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

Determine the decay constant of Banfordium.

Possible Answers:

$\frac{.525}{year}$ $\displaystyle \frac{.525}{year}$

$\frac{.825}{year}$ $\displaystyle \frac{.825}{year}$

None of these

$\frac{.711}{year}$ $\displaystyle \frac{.711}{year}$

$\frac{.448}{year}$ $\displaystyle \frac{.448}{year}$

Correct answer:

$\frac{.525}{year}$ $\displaystyle \frac{.525}{year}$

Explanation:

Calculate the decay constant:

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

Solve for $\lambda$ $\displaystyle \lambda$

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

Plug in values:

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

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

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Example Question #61 : Quantum And Nuclear Physics

Determine the half-life of Banfordium.

Possible Answers:

$t_{\frac{1}{2}}=10yr$ $\displaystyle t_{\frac{1}{2}}=10yr$

$t_{\frac{1}{2}}=1.32yr$ $\displaystyle t_{\frac{1}{2}}=1.32yr$

$t_{\frac{1}{2}}=30yr$ $\displaystyle t_{\frac{1}{2}}=30yr$

$t_{\frac{1}{2}}=2.72yr$ $\displaystyle t_{\frac{1}{2}}=2.72yr$

$t_{\frac{1}{2}}=7.8yr$ $\displaystyle t_{\frac{1}{2}}=7.8yr$

Correct answer:

$t_{\frac{1}{2}}=1.32yr$ $\displaystyle t_{\frac{1}{2}}=1.32yr$

Explanation:

Calculate the decay constant:

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

Solve for $\lambda$ $\displaystyle \lambda$

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

Plug in values:

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

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

Use the following relationship:

$t_{\frac{1}{2}}=\frac{ln2}{\lambda}$ $\displaystyle t_{\frac{1}{2}}=\frac{ln2}{\lambda}$

Plugging in values

$t_{\frac{1}{2}}=\frac{ln2}{.525}$ $\displaystyle t_{\frac{1}{2}}=\frac{ln2}{.525}$

$t_{\frac{1}{2}}=1.32yr$ $\displaystyle t_{\frac{1}{2}}=1.32yr$

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Example Question #391 : Ap Physics 2

Determine the number of Banfordium nuclei in a sample of activity $3.1*10^{13}Bq$ $\displaystyle 3.1*10^{13}Bq$

Possible Answers:

$3.41*10^{18}nuclei=N$ $\displaystyle 3.41*10^{18}nuclei=N$

$8.55*10^{19}nuclei=N$ $\displaystyle 8.55*10^{19}nuclei=N$

$6.72*10^{18}nuclei=N$ $\displaystyle 6.72*10^{18}nuclei=N$

$7.77*10^{19}nuclei=N$ $\displaystyle 7.77*10^{19}nuclei=N$

$1.15*10^{20}nuclei=N$ $\displaystyle 1.15*10^{20}nuclei=N$

Correct answer:

$7.77*10^{19}nuclei=N$ $\displaystyle 7.77*10^{19}nuclei=N$

Explanation:

Calculate the decay constant:

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

Solve for $\lambda$ $\displaystyle \lambda$

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

Plug in values

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

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

Convert to $s^{-1}$ $\displaystyle s^{-1}$

$\frac{.525}{year}*\frac{1 year}{365.25days}*\frac{1 day}{3600 s}=\frac{3.99*10^{-7}}{s}$ $\displaystyle \frac{.525}{year}*\frac{1 year}{365.25days}*\frac{1 day}{3600 s}=\frac{3.99*10^{-7}}{s}$

Use the following relationship:

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

Plug in values and solve:

$3.1*10^{13}=3.99*10^{-7}*N$ $\displaystyle 3.1*10^{13}=3.99*10^{-7}*N$

$7.77*10^{19}nuclei=N$ $\displaystyle 7.77*10^{19}nuclei=N$

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Example Question #41 : Atomic And Nuclear Physics

Determine the activity of the sample 10 years after the initial test.

Possible Answers:

$2.2*10^{11}Bq$ $\displaystyle 2.2*10^{11}Bq$

$5.1*10^{11}Bq$ $\displaystyle 5.1*10^{11}Bq$

$4.3*10^{11}Bq$ $\displaystyle 4.3*10^{11}Bq$

$1.7*10^{11}Bq$ $\displaystyle 1.7*10^{11}Bq$

$8.5*10^{11}Bq$ $\displaystyle 8.5*10^{11}Bq$

Correct answer:

$2.2*10^{11}Bq$ $\displaystyle 2.2*10^{11}Bq$

Explanation:

Calculate the decay constant

Using

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

Solve for $\lambda$ $\displaystyle \lambda$

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

Plug in values:

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

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

Use the following relationship:

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

Plug in values:

$A=4.2*10^{13}e^{-.525*10}$ $\displaystyle A=4.2*10^{13}e^{-.525*10}$

$A=2.2*10^{11}Bq$ $\displaystyle A=2.2*10^{11}Bq$

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Example Question #62 : Quantum And Nuclear Physics

A sample of radioactive iodine is found to have decreased over time. After $\displaystyle 4$ days, $18\:mg$ $\displaystyle 18\:mg$ of radioactive iodine is found. Given that the half-life of radioactive iodine is 8.05 $\displaystyle 8.05$ days, how much radioactive iodine was there initially?

Possible Answers:

$71.8\:mg$ $\displaystyle 71.8\:mg$

$97.4\:mg$ $\displaystyle 97.4\:mg$

$32.2\:mg$ $\displaystyle 32.2\:mg$

$48.3\:mg$ $\displaystyle 48.3\:mg$

$25.4\:mg$ $\displaystyle 25.4\:mg$

Correct answer:

$25.4\:mg$ $\displaystyle 25.4\:mg$

Explanation:

In this question, we're presented with a case in which a certain amount of radioactive iodine has decayed in a given amount of time. Based on the half-life, the amount of time that has passed, and the current amount of radioactive iodine, we're asked to solve for the initial amount of iodine.

First off, it's important to realize that this is a radioactive decay problem. All radioactive decay processes occur by a first-order reaction mechanism. Thus, we can go ahead and use the equation for first-order reactions:

$A_{t}=A_{0}e^{-kt}$ $\displaystyle A_{t}=A_{0}e^{-kt}$

Rearranging to isolate the $A_{0}$ $\displaystyle A_{0}$ term, we obtain:

$A_{0}=A_{t}e^{kt}$ $\displaystyle A_{0}=A_{t}e^{kt}$

But before we start plugging in values, we'll need to calculate the $\displaystyle k$ term, which is the reaction rate constant. We can do this because we know that the reaction is first-order, and we're also provided with the half-life. Using the rate constant equation for a first-order reaction:

$k=-\frac{ln(\frac{1}{2})}{t_{\frac{1}{2}}}=\frac{0.693}{8.05\: days}=0.086\:days^{-1}$ $\displaystyle k=-\frac{ln(\frac{1}{2})}{t_{\frac{1}{2}}}=\frac{0.693}{8.05\: days}=0.086\:days^{-1}$

Next, we can plug in the values given to us in order to solve for our answer:

$A_{0}=(18\: mg)e^{(0.086\: days^{-1})(4\: days)}$ $\displaystyle A_{0}=(18\: mg)e^{(0.086\: days^{-1})(4\: days)}$

$A_{0}=25.4\:mg$ $\displaystyle A_{0}=25.4\:mg$

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Example Question #41 : Atomic And Nuclear Physics

The half life of $P^{238}$ $\displaystyle P^{238}$ is $\displaystyle 88 years$. Determine the radioactive decay constant.

Possible Answers:

$\frac{.712}{year}$ $\displaystyle \frac{.712}{year}$

$\frac{.00765}{year}$ $\displaystyle \frac{.00765}{year}$

$\frac{.458}{year}$ $\displaystyle \frac{.458}{year}$

$\frac{.00785}{year}$ $\displaystyle \frac{.00785}{year}$

$\frac{.00115}{year}$ $\displaystyle \frac{.00115}{year}$

Correct answer:

$\frac{.00785}{year}$ $\displaystyle \frac{.00785}{year}$

Explanation:

Use the following formula:

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

Plug in values:

$88years=\frac{.693}{\lambda}$ $\displaystyle 88years=\frac{.693}{\lambda}$

$\lambda=\frac{.00785}{year}$ $\displaystyle \lambda=\frac{.00785}{year}$

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Example Question #41 : Atomic And Nuclear Physics

The half life of $P^{238}$ $\displaystyle P^{238}$ is $\displaystyle 88 years$. Determine the number of radioactive nuclei in a sample with an activity of 750Bq $\displaystyle 750Bq$.

Possible Answers:

$7.23*10^{11}$ $\displaystyle 7.23*10^{11}$

$3.91*10^{11}$ $\displaystyle 3.91*10^{11}$

$5.55*10^{11}$ $\displaystyle 5.55*10^{11}$

$1.26*10^{11}$ $\displaystyle 1.26*10^{11}$

$3.33*10^{11}$ $\displaystyle 3.33*10^{11}$

Correct answer:

$1.26*10^{11}$ $\displaystyle 1.26*10^{11}$

Explanation:

Use the following formula:

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

Plug in values:

$88years=\frac{.693}{\lambda}$ $\displaystyle 88years=\frac{.693}{\lambda}$

$\lambda=\frac{.00785}{year}$ $\displaystyle \lambda=\frac{.00785}{year}$

Covert to $s^{-1}$ $\displaystyle s^{-1}$

$\frac{.00785}{year}*\frac{1 year}{365.25 days}*\frac{1 day}{3600 s}=\frac{5.97*10^{-9}}{s}$ $\displaystyle \frac{.00785}{year}*\frac{1 year}{365.25 days}*\frac{1 day}{3600 s}=\frac{5.97*10^{-9}}{s}$

Use the relationship:

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

Plug in values:

$750=5.97*10^{-9} *N$ $\displaystyle 750=5.97*10^{-9} *N$

$N=1.26*10^{11}$ $\displaystyle N=1.26*10^{11}$

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Example Question #41 : Atomic And Nuclear Physics

A sample is taken of a radioactive element. It has an activity of 1991 Bq $\displaystyle 1991 Bq$. Two hours later, it has an activity of 28 Bq $\displaystyle 28 Bq$.

Determine the nuclear decay constant in terms of $min^{-1}$ $\displaystyle min^{-1}$.

Possible Answers:

$\frac{.0671}{min}$ $\displaystyle \frac{.0671}{min}$

None of these

$\frac{.0355}{min}$ $\displaystyle \frac{.0355}{min}$

$\frac{.0539}{min}$ $\displaystyle \frac{.0539}{min}$

$\frac{.0222}{min}$ $\displaystyle \frac{.0222}{min}$

Correct answer:

$\frac{.0355}{min}$ $\displaystyle \frac{.0355}{min}$

Explanation:

Determination of decay constant:

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

Solving for $\lambda$ $\displaystyle \lambda$

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

Converting hours to minutes and plugging in values:

$-ln(\frac{28}{1991})*\frac{1}{120}=\lambda$ $\displaystyle -ln(\frac{28}{1991})*\frac{1}{120}=\lambda$

$\frac{.0355}{min}=\lambda$ $\displaystyle \frac{.0355}{min}=\lambda$

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Example Question #42 : Atomic And Nuclear Physics

A sample is taken of a radioactive element. It has an activity of 1991 Bq $\displaystyle 1991 Bq$. Two hours later, it has an activity of 28 Bq $\displaystyle 28 Bq$.

Determine the half life.

Possible Answers:

27.5min $\displaystyle 27.5min$

32.4min $\displaystyle 32.4min$

None of these

19.5min $\displaystyle 19.5min$

12.3min $\displaystyle 12.3min$

Correct answer:

19.5min $\displaystyle 19.5min$

Explanation:

Determination of decay constant:

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

Solving for $\lambda$ $\displaystyle \lambda$

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

Converting hours to minutes and plugging in values:

$-ln(\frac{28}{1991})*\frac{1}{120}=\lambda$ $\displaystyle -ln(\frac{28}{1991})*\frac{1}{120}=\lambda$

$\frac{.0355}{min}=\lambda$ $\displaystyle \frac{.0355}{min}=\lambda$

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

$t_{1/2}=\frac{ln(2)}{.0355}$ $\displaystyle t_{1/2}=\frac{ln(2)}{.0355}$

$t_{1/2}=19.5min$ $\displaystyle t_{1/2}=19.5min$

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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 #21 : Radioactive Nuclear Decay

Example Question #391 : Ap Physics 2

Example Question #61 : Quantum And Nuclear Physics

Example Question #391 : Ap Physics 2

Example Question #41 : Atomic And Nuclear Physics

Example Question #62 : Quantum And Nuclear Physics

Example Question #41 : Atomic And Nuclear Physics

Example Question #41 : Atomic And Nuclear Physics

Example Question #41 : Atomic And Nuclear Physics

Example Question #42 : Atomic And Nuclear Physics

All AP Physics 2 Resources

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