Question

The radioactive disintegration of ${}_{94}^{239}\text{Pu}$, an $\alpha$-emission process is accompanied by the loss of 5.24 MeV/dis. If $t_{1/2}$ of ${}_{94}^{239}\text{Pu}$ is $2.44\times {10}^4$year, calculate the energy released per year from 1.0 g sample of ${}_{94}^{239}\text{Pu}$ in kJ is : (give answer in X/10 form)

Answer 1

$k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}log\frac{a}{a-x}$

Substitute values in the above equation.

$\frac{0.693}{2.44\times {10}^4}=\frac{2.303}{1}log\frac{1}{a-x}$

$1.233\times {10}^{-5}=log\frac{1}{a-x}$

$a-x=0.999972$

$x=1-0.999972=2.839\times {10}^{-5}$

Hence, the number of disintegrations are $\frac{2.839\times {10}^{-5}}{239}\times 6.023\times {10}^{23}=7.155\times {10}^{16}$.

The energy released is $5.24MeV/dis\times 7.155\times {10}^{16}dis\times 1.602\times {10}^{-16}kJ/mol=60kJ$.

So, energy released is $\frac{60}{10}=6$.