AP Physics 2 - Radioactive Nuclear Decay

Suppose that the half-life of substance A is known to be . When substance A decays, it becomes substance B. If this is the only way that substance B can be produced, and a sample is found that contains of substance A and of substance B, what is the age of this sample?

Explanation

To find the approximate age of the sample, we need to consider the half-life of the starting material, substance A. We also need to consider the relative amounts of substance A and substance B.

Since we are told that the sample we are looking at contains substance A and substance B, we know that the sample must have started out with of substance A. Furthermore, in order to go from to , we know that a total of two half-lives must have passed. And since we know that one half-life is equal to , we can conclude that a total of must have passed.

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?

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.

$\left(\frac{1}{2} \right )^{0.039267}=0.973\cdot 100%=97.3%$

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.

None of these

There will be no activity left

Explanation

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.

$\frac{(-ln(\frac{2.9*10^4 Bq}{3.3*10^4 Bq}))}{17years}=\lambda$

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

Again using the relationship

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

Using the new

$A=3.3*10^{4}e^{-{7.6*10^{-3} 23}}$

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.

None of these

Explanation

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.

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

$\frac{3.83*10^{-3}}{s}=\lambda$

Using the relationship

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

Plugging in the calculated value for $\lambda$

$\frac{ln(2))}{3.83*10^{-3}}=t_{1/2}$

$181s=t_{1/2}$

A scientist runs a test on a radioactive sample and finds that it has an activity of . Three days later, she runs another test and finds the activity to be .

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

None of these

Explanation

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 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

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

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

$1.26*10^{11}$

$3.91*10^{11}$

$3.33*10^{11}$

$5.55*10^{11}$

$7.23*10^{11}$

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:

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

$N=1.26*10^{11}$

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.

None of these

Explanation

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.

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

$\frac{9.86*10^{-4}}{s}=\lambda$

Again use the relationship:

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

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

$A=1868e^{-9.86*10^{-4} 1080}$

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 $8.78*10^{12} Bq$ .