Chapter 30, Problem 71A

To determine

To Calculate: The fraction of a sample that would be left for the isotope $\text{R}\text{n}$ .

Answer to Problem 71A

  $N(25)=0.347$

Explanation of Solution

Given:

Half-life= $15\text{ h}$

Fraction of sample would be left after 60 h =?

Formula used:

The fraction of molecules $N(t)$ after t half lives depends exponentially as:

  $N(t)=N_0{e}^{\left(-\lambda .t\right)}$

Where,

  $N_0$ is initial number

  $t$ is number of half lives

Calculation:

Consider the table.

Time	half live	fraction left
0	0	$100\%$
$15\text{ hours}$	1	$50\%$
$60\text{ hours}$	2	$25\%$

The mathematical solution is,

  $\lambda =\frac{0.693}{T_{\text{half}}}$

  $T_{\text{half}}=15\text{ hours}$

  $\lambda =0.0462$

  $N_0=1$

  $N(t)=N_0{e}^{\left(-\lambda .t\right)}$

  $t=60$

  $N(25)=1\times e^{\left(-0.0423\times 25\right)}$

  $N(25)=0.347$

The remaining fraction is $\mathrm{0.347.}$

Conclusion:

The fraction left after the $60\text{ hours}$ is $\mathrm{0.347.}$