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

Possible Answers:

$5.0 \times 10^{23}atoms$

$1.25 \times 10^{23}atoms$

$10.0 \times 10^{23}atoms$

$25.0 \times 10^{23}atoms$

$2.5 \times 10^{23}atoms$

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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Example Question #31 : Radioactive Nuclear Decay

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 half-life of Banfordium.

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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

Use the following relationship:

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

Plugging in values

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

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

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Example Question #61 : Quantum And Nuclear Physics

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

Convert to $s^{-1}$

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

Use the following relationship:

$A=\lambda*N$

Plug in values and solve:

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

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

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Example Question #41 : Atomic And Nuclear Physics

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 activity of the sample 10 years after the initial test.

Possible Answers:

$2.2*10^{11}Bq$

$5.1*10^{11}Bq$

$4.3*10^{11}Bq$

$1.7*10^{11}Bq$

$8.5*10^{11}Bq$

Correct answer:

$2.2*10^{11}Bq$

Explanation:

Calculate the decay constant

Using

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

Solve for $\lambda$

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

Plug in values:

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

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

Use the following relationship:

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

Plug in values:

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

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

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Example Question #62 : Quantum And Nuclear Physics

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

Possible Answers:

$32.2\:mg$

$25.4\:mg$

$48.3\:mg$

$71.8\:mg$

$97.4\:mg$

Correct answer:

$25.4\:mg$

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:

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

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

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

But before we start plugging in values, we'll need to calculate the 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:

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

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

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

$A_{0}=25.4\:mg$

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Example Question #41 : Atomic And Nuclear Physics

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

Possible Answers:

$\frac{.712}{year}$

$\frac{.00765}{year}$

$\frac{.458}{year}$

$\frac{.00785}{year}$

$\frac{.00115}{year}$

Correct answer:

$\frac{.00785}{year}$

Explanation:

Use the following formula:

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

Plug in values:

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

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

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Example Question #41 : Atomic And Nuclear Physics

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

Possible Answers:

$1.26*10^{11}$

$5.55*10^{11}$

$7.23*10^{11}$

$3.91*10^{11}$

$3.33*10^{11}$

Correct answer:

$1.26*10^{11}$

Explanation:

Use the following formula:

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

Plug in values:

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

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

Covert to $s^{-1}$

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

Use the relationship:

$A=\lambda N$

Plug in values:

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

$N=1.26*10^{11}$

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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 . Two hours later, it has an activity of 28 Bq .

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

Possible Answers:

$\frac{.0671}{min}$

None of these

$\frac{.0355}{min}$

$\frac{.0539}{min}$

$\frac{.0222}{min}$

Correct answer:

$\frac{.0355}{min}$

Explanation:

Determination of decay constant:

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

Solving for $\lambda$

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

Converting hours to minutes and plugging in values:

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

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

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Example Question #61 : Quantum And Nuclear Physics

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

Determine the half life.

Possible Answers:

None of these

27.5min

12.3min

32.4min

19.5min

Correct answer:

19.5min

Explanation:

Determination of decay constant:

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

Solving for $\lambda$

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

Converting hours to minutes and plugging in values:

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

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

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

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

$t_{1/2}=19.5min$

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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 #392 : Ap Physics 2

Example Question #31 : Radioactive Nuclear Decay

Example Question #61 : Quantum And Nuclear Physics

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 #61 : Quantum And Nuclear Physics

All AP Physics 2 Resources

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