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AP Physics 2 : AP Physics 2

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Example Questions

Example Question #21 : Quantum And Nuclear Physics

Which of the following particles has a charge that is fractions of an electron charge?

Possible Answers:

Hadron

Tau

Quark

Graviton

Tachyon

Correct answer:

Quark

Explanation:

The correct answer is quarks. Quarks usually have charges of $\pm\frac{1}{3}e$ or $\pm\frac{2}{3}e$ . They are usually bound with other quark particles and could be mixed to form hadrons. Tau is part of the leptons family and has a charge of -1e . Graviton does not have a charge, and is a hypothetical particle. The tachyon is a hypothetical particle assumed to be faster than light. Hadrons are strong composite particles that are composed of quarks and will result to a net integer charge.

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Example Question #361 : Ap Physics 2

Which of the following subatomic particles has the highest charge to mass ratio?

Possible Answers:

Neutron

Gluon

Electron

Proton

Antiproton

Correct answer:

Electron

Explanation:

Neutrons and gluons both have no charge at all, so they can be ignored. The proton and the antiproton have the same mass but opposite charges, and so have the same ration of charge:mass. However, the electron has equal charge to both the proton and the antiproton, and has a ridiculously small mass comparatively. The mass of a proton/antiproton is $1.67262178 * 10^{-27} \textup{ kg}$ , while the mass of an electron is $9.10938291 * 10^{-31} \textup{ kg}$ , almost 2000 times as small as a proton. Therefore, the charge:mass ratio of the electron is the smallest of the fundamental particles listed.

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Example Question #361 : Ap Physics 2

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:

$75\%$

$64\%$

$50\%$

$25\%$

$36\%$

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 #362 : Ap Physics 2

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

Possible Answers:

None of these particles are known as positrons

$\beta^+$ particle

$\gamma$ ray

$\beta^-$ particle

$\alpha$ 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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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 : Quantum And Nuclear Physics

Example Question #361 : Ap Physics 2

Example Question #361 : Ap Physics 2

Example Question #362 : Ap Physics 2

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

All AP Physics 2 Resources

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