Chapter 30, Problem 7STP

To determine

To Choose: The correct option.

Answer to Problem 7STP

C

Explanation of Solution

Given:

Half-life of polonium- $210$

  $=138days$

Starting with $2.34kg$ sample

Formula used:

  $A=A_0\times \frac{1}{2_n}$

Where,

  $A$ is amount of sample after $n$ half lives

  $A_0$ is the initial amount

Calculation:

The time needed for an initial sample of a polonium- $210$ to be halved.

Start with A grams of polonium- $210$ left with,

One half-life - $A\times \frac{1}{2}=\frac{A}{2}$

Two half-life- $A\times \frac{1}{2^2}=\frac{A}{4}$

Three half-life- $A\times \frac{1}{2^3}=\frac{A}{8}$

Now use the formula, to determine the remain polonium- $210$

  $A=A_0\times \frac{1}{2^n}$ -----------1

Where A₀ = Initial mass of sample

  $n=\frac{\text{4×365 days}}{\text{138 days}}$

  $n=10.57$

Substitute the values in the equation 1

  $A=2.32\text{ kg}\times \frac{1}{2^{10.57}}$$$

  $A=3.2\text{ kg}\times \frac{1}{2^{10.57}}$

  $\begin{array}{l}A=1.53\times {10}^{-3}\text{ kg}\\ A=1.53\text{ grams}\end{array}$

Conclusion:

  $1.53grams$ of a $2.3kg$ sample will remain after four years.