Chapter 30, Problem 4STP

To determine

To Choose: The correct option.

Answer to Problem 4STP

Option B

Explanation of Solution

Given info:

half-life of iodine-131, $t_{1/2}=8\text{days}$

Time, $t=16\text{days}$

Initial activity, $A_0=2.5\times {10}^8\text{Bq}$

Formula used:

• The number of half-lives is calculated by the formula

  $n=\frac{\text{time}}{\text{half-lives}}=\frac{t}{t_{1/2}}$

• The activity if the element Inodine-131 after 16 days is calculated as

  $A_t=A_0\times \frac{1}{2^n}$

Where, n is the number of half-lives, $A_0$ is the initial activity, and $A_t$ is the activity at any time “t”.

Calculation:

Firstly, calculate the number of half-lives by substituting the values of time such that

  $n=\frac{t}{t_{1/2}}=\frac{\text{16days}}{\text{8days}}=2$

For $n=2$ , here the activity of the given sample after 16 days will be

  $\begin{array}{l}{A}_t=A_0\times \frac{1}{2^n}=2.5\times {10}^8\text{Bq}\times \frac{1}{2^2}\\ =2.5\times {10}^8\text{Bq}\times \frac{1}{4}=0.625\times {10}^8\text{Bq}\\ A_t=6.3\times {10}^7\text{Bq}\end{array}$

Conclusion:

The activity is $6.2\times {10}^7\text{Bq}$ .