Quantum and Nuclear Physics

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AP Physics 2 › Quantum and Nuclear Physics

Questions 1 - 10

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.

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

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.

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

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.

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

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.

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.

A long rod is traveling at in relationship to an observer along it's long axis. Determine the observed length.

None of these

Explanation

Using

$L=L_0\sqrt{1-\frac{v^2}{c^2}}$

Plugging in values

$L=3*\sqrt{1-\frac{(.75c)^2}{c^2}}$

Determine the observed length of a $7\textup{ m}$ rod traveling along it's long axis at $0.86\textup{ c}$ in relation to an observer.

$3.57\textup{ m}$

None of these

$6.68\textup{ m}$

$2.54\textup{ m}$

$8.44\textup{ m}$

Explanation

Use the following equation:

$L=L_0\sqrt{1-\frac{v^2}{c^2}}$

Where is the rest length,

is the velocity of the object

is the speed of light

is the observed length

Plugging in values

$L=7\sqrt{1-\frac{(.86c)^2}{c^2}}$

$L=3.57\textup{ m}$

Determine the observed length of a $7\textup{ m}$ rod traveling along it's long axis at $0.86\textup{ c}$ in relation to an observer.

$3.57\textup{ m}$

None of these

$6.68\textup{ m}$

$2.54\textup{ m}$

$8.44\textup{ m}$

Explanation

Use the following equation:

$L=L_0\sqrt{1-\frac{v^2}{c^2}}$

Where is the rest length,

is the velocity of the object

is the speed of light

is the observed length

Plugging in values

$L=7\sqrt{1-\frac{(.86c)^2}{c^2}}$

$L=3.57\textup{ m}$

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.