A 6.25mg sample of Cr51 decays for 111 days. After that amount of time, 0.75mg remains. What is the half-life of Cr51?

Answers

Answer 1

Answer: To find the half-life (\( t_{\text{half}} \)) of Cr51, we can use the radioactive decay formula:

\[ N = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{\text{half}}}} \]

Where:
- \( N \) is the remaining amount after time \( t \) (0.75 mg in this case)
- \( N_0 \) is the initial amount (6.25 mg)
- \( t \) is the time elapsed (111 days in this case)
- \( t_{\text{half}} \) is the half-life we're trying to find

We'll rearrange this formula to solve for \( t_{\text{half}} \):

\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{t_{\text{half}}}} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{N}{N_0}\right) = \frac{-t}{t_{\text{half}}} \times \ln\left(\frac{1}{2}\right) \]

Now, plug in the given values:

\[ \ln\left(\frac{0.75\, \text{mg}}{6.25\, \text{mg}}\right) = \frac{-111\, \text{days}}{t_{\text{half}}} \times \ln\left(\frac{1}{2}\right) \]

Solve for \( t_{\text{half}} \):

\[ t_{\text{half}} = \frac{-111\, \text{days}}{\ln\left(\frac{0.75\, \text{mg}}{6.25\, \text{mg}}\right) / \ln\left(\frac{1}{2}\right)} \]

Now, calculate this to find the half-life.

Using the provided formula to calculate the half-life (\( t_{\text{half}} \)), we get:

\[ t_{\text{half}} = \frac{-111\, \text{days}}{\ln\left(\frac{0.75\, \text{mg}}{6.25\, \text{mg}}\right) / \ln\left(\frac{1}{2}\right)} \]

Plugging in the values and calculating:

\[ t_{\text{half}} ≈ \frac{-111\, \text{days}}{-1.4978} \]
\[ t_{\text{half}} ≈ 74.24\, \text{days} \]

So, the half-life of Cr51 is approximately \( 74.24\, \text{days} \).

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if you need more explanation , comment.

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Hope this helps!

Hey there

The correct answer is (C)

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