Chapter 7.4, Problem 10E

a.

To determine

To calculate : the half life of tritium- $3$

Answer to Problem 10E

The half life of tritium- $3$ is $12.25$

Explanation of Solution

Given information : The question is

A sample of tritium- $3$ decayed to $94.5\%$ of its original amount after a year

Calculation : The half life of tritium- $3$ is

  $A=A_0{e}^{-kt}$

  $t$ is inn year

  $0.945=e^{-k}\mathrm{......}\left(1\right)$

Let half life be $x$ then,

  $\begin{array}{l}A=0.5A_0\\ 0.5=e^{-kx}\\ 0.5={\left(e^{-k}\right)}^x\\ 0.5={\left(0.945\right)}^x\\ \ln \left(0.5\right)=x\ln \left(0.945\right)\\ x=\frac{\ln \left(0.5\right)}{\ln \left(0.945\right)}=12.25\end{array}$

So, the half life of tritium- $3$ is $12.25$

b.

To determine

To calculate : that how long the sample to decay of $20\%$ of its original amount

Answer to Problem 10E

It takes $28.45years$ to decay of $20\%$ of its original amount

Explanation of Solution

Given information : The question is

A sample of tritium- $3$ decayed to $94.5\%$ of its original amount after a year

Calculation : If $t_1$ is the time it takes the sample to decay to $20\%$ of its original amount

  $\begin{array}{l}0.2=N\left(t_1\right)={\left(0.945\right)}^{t_1}\\ \ln \left(0.2\right)=\ln \left({0.945}^{t_1}\right)=t_1\ln \left(0.945\right)\\ t_1=\frac{\ln \left(0.2\right)}{\ln \left(0.945\right)}\approx 28.45years\end{array}$