Question

1. calculate each of the following involving half - lives:

a. the half - life of na - 24 is 15 h. if the initial activity of the sodium - 24 is 6.4 mci, what is its activity after 60. h.
b. if a sample of i - 123 with an activity of 4 kbq decays to an activity of 1 kbq in 26.4 h, what is its half - life?

Explanation:

Step1: Calculate the number of half - lives for part a

The half - life of \(Na - 24\) is \(T = 15h\), and the time elapsed is \(t=60h\). The number of half - lives \(n=\frac{t}{T}\).

$$n=\frac{60}{15}=4$$

Step2: Use the formula for radioactive decay for part a

The formula for radioactive decay is \(A = A_{0}(\frac{1}{2})^{n}\), where \(A_{0}\) is the initial activity and \(A\) is the final activity. Given \(A_{0}=6.4mCi\) and \(n = 4\).

$$A=6.4\times(\frac{1}{2})^{4}$$

$$A = 6.4\times\frac{1}{16}$$

$$A=0.4mCi$$

Step3: Calculate the number of half - lives for part b

Let the half - life be \(T\). The initial activity \(A_{0}=4kBq\), the final activity \(A = 1kBq\), and the time elapsed \(t = 26.4h\). Using the formula \(A = A_{0}(\frac{1}{2})^{n}\), we can find \(n\) first.

$$1=4\times(\frac{1}{2})^{n}$$

$$\frac{1}{4}=(\frac{1}{2})^{n}$$

, so \(n = 2\)

Step4: Calculate the half - life for part b

Since \(n=\frac{t}{T}\), and \(n = 2\), \(t = 26.4h\)

$$T=\frac{t}{n}=\frac{26.4}{2}=13.2h$$

Answer:

a. \(0.4mCi\)
b. \(13.2h\)