Radioactive Nuclear Decay - AP Physics 2

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Question

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?

Answer

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 as much of Compound X as there was when at the beginning. Since there is remaining after this time period, then we can conclude that of Compound X has degraded within this amount of time.

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Question

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

Answer

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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Question

A test is done on a sample of a newly discovered radioactive nuclei, which has an activity of . later, it has an activity of .

Determine the activity after the initial reading.

Answer

Using the relationship:

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

Here, is the activity at a given time, 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.

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

$\frac{3.8310^{-3}}{s}=\lambda$

Again using the relationship

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

Using the new , which is equal to

$A=1990e^{-3.8310^{-3} 1080}$

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Question

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?

Answer

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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Question

You measure the beta decay activity of an unknown substance to be . 48 hours later, the activity is .

What is the half life in hours?

Answer

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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Question

A particular sample of a newly discovered isotope has an activity of . 10 minutes later, it has an activity of .

Determine the radioactive decay constant of this isotope.

Answer

Use the relationship:

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

Here, is the activity at a given time, 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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Question

A scientist tests a radioactive sample which has an activity of . 15 minutes later, it has an activity of .

Determine the number of radioactive nuclei in the initial sample.

Answer

Use the relationship:

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

Where is the activity at a given time, 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.

$\frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\frac{9.8610^{-4}}{s}=\lambda$

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:

$1868= \frac{9.8410^{-4}}{s} N$

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Question

A scientist tests a radioactive sample which has an activity of . 15 minutes later, it has an activity of .

Determine the activity 18 minutes after the initial reading.

Answer

Use the relationship:

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

Where is the activity at a given time, 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.

$\frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\frac{9.8610^{-4}}{s}=\lambda$

Again use the relationship:

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

Use the new , which is equal to to plug in and solve for the activity.

$A=1868e^{-9.8610^{-4} 1080}$

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Question

A scientist test a radioactive sample which has an activity of . 15 minutes later, it has an activity of .

Determine the half life of this isotope.

Answer

Use the relationship:

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

Here, is the activity at a given time, 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.

$\frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\frac{9.8610^{-4}}{s}=\lambda$

Use the relationship:

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

Plug in the calculated value for $\lambda$ :

$\frac{ln(2))}{9.8610^{-4}}=t_{\frac{1}{2}}$

$702s=t_{\frac{1}{2}}$

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Question

A scientist test a radioactive sample which has an activity of . 15 minutes later, it has an activity of .

Determine the nuclear decay constant of this isotope.

Answer

Use the relationship:

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

Here, is the activity at a given time, 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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Question

A scientist takes a reading of a radioactive material, which has an activity of . 15 minutes later, it has an activity of .

Determine the number of radioactive atoms in the initial sample.

Answer

Use the relationship:

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

Where is the activity at a given time, 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.

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

$\frac{5.8810^{-4}}{s}=\lambda$

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:

$1570= \frac{5.8810^{-4}}{s} N$

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Question

A scientist takes a reading of a radioactive material, which has an activity of . 15 minutes later, it has an activity of .

Determine the activity at 25 minutes after the initial reading.

Answer

Use the relationship:

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

Here, is the activity at a given time, 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.

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

$\frac{5.8810^{-4}}{s}=\lambda$

Again use the relationship:

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

Using the new , which is equal to

$A=1570e^{-5.8810^{-4} 1500}$

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Question

A scientist takes a reading of a radioactive material, which has an activity of . 15 minutes later, it has an activity of .

Determine the half-life of this isotope.

Answer

Use the relationship:

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

Here, is the activity at a given time, 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.

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

$\frac{5.8810^{-4}}{s}=\lambda$

Use the relationship:

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

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

$\frac{ln(2))}{5.8810^{-4}}=t_{\frac{1}{2}}$

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

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Question

A scientist takes a reading of a radioactive material, which has an activity of . 15 minutes later, it has an activity of .

Determine the nuclear decay constant of this isotope.

Answer

Use the relationship:

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

Here, is the activity at a given time, 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.

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

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

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Question

A test is done on a sample of a newly discovered radioactive nuclei, which has an activity of . later, it has an activity of .

Determine the half life of this nuclei.

Answer

Using the relationship:

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

Here, is the activity at a given time, 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.

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

$\frac{3.8310^{-3}}{s}=\lambda$

Using the relationship

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

Plugging in the calculated value for $\lambda$

$\frac{ln(2))}{3.8310^{-3}}=t_{1/2}$

$181s=t_{1/2}$

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Question

A scientist discovers a new radioactive nuclei. She runs a test on a sample and finds it has an activity of . later, it has an activity of .

Determine the decay constant.

Answer

Using the relationship:

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

Here, is the activity at a given time, 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.

$\frac{(-ln(\frac{2.910^4 Bq}{3.310^4 Bq}))}{17years}=\lambda$

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

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Question

A scientist discovers a new radioactive nuclei. She runs a test on a sample and finds it has an activity of . later, it has an activity of .

Determine the half life.

Answer

Using the relationship:

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

Here, is the activity at a given time, 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.

$\frac{(-ln(\frac{2.910^4 Bq}{3.310^4 Bq}))}{17years}=\lambda$

$\frac{7.610^{-3}}{year}=\lambda$

Using the relationship

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

Plugging in the calculated value for $\lambda$

$\frac{ln(2))}{7.610^{-3}}=t_{1/2}$

$91.2years=t_{1/2}$

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Question

A scientist discovers a new radioactive nuclei. She runs a test on a sample and finds it has an activity of . later, it has an activity of .

Determine the number of radioactive atoms in the initial sample.

Answer

Using the relationship:

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

Here, is the activity at a given time, 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.

$\frac{(-ln(\frac{2.910^4 Bq}{3.310^4 Bq}))}{17years}=\lambda$

$\frac{7.610^{-3}}{year}=\lambda$

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.

$\frac{7.610^{-3}}{year}\frac{1year}{365.25days}\frac{1day}{3600seconds}=\frac{5.7810^{-9}}{second}$

Using the initial activity and the calculated decay constant:

$3.310^{4}= \frac{5.7810^{-9}}{s} N$

$N=5.7110^{12}$

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Question

A scientist discovers a new radioactive nuclei. She runs a test on a sample and finds it has an activity of . later, it has an activity of .

Determine the activity after the original reading.

Answer

Using the relationship:

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

Here, is the activity at a given time, 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.

$\frac{(-ln(\frac{2.910^4 Bq}{3.310^4 Bq}))}{17years}=\lambda$

$\frac{7.610^{-3}}{year}=\lambda$

Again using the relationship

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

Using the new

$A=3.310^{4}e^{-{7.6*10^{-3} 23}}$

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Question

A scientist takes a sample of a newly discovered radioactive element, which has an activity of . 10 minutes later, it has an activity of .

Determine the number of radioactive nuclei in the original sample.

Answer

Use the relationship:

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

Where is the activity at a given time, 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.

$\frac{(-ln(\frac{220 Bq}{1990 Bq}))}{600s}=\lambda$

$\frac{3.6710^{-3}}{s}=\lambda$

It is then necessary to use the relationship:

$A= \lambda N$

Use the initial activity and the calculated decay constant:

$1990= \frac{3.6710^{-3}}{s} N$

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