Question

A star initially has ${10}^{40}$ deutrons. It produces energy via the processes
$\begin{array}{c}{H}^2+,H^2⟶,H^3+p\\ H^2+H^3{⟶}_{2}H{e}^4+n\end{array}$

The masses of the nuclei are as follows$\begin{array}{c}M\left(H^2\right)=2.014\text{amu:}M\left(p\right)=1.007\text{amu:}\\ M\left(n\right)=1.008\text{amu;}M\left(H{e}^4\right)=4.00\text{lamu}\end{array}$

If the average power radiated by the star is ${10}^{16}$ w,the deutron supply of the star is exhausted in a time of the order of

• A. ${10}^{6}s$
• B. ${10}^{8}s$
• C. ${10}^{12}$
• D. ${10}^{16}s$

Answer 1

Answer: (C)

${10}^{12}$

$\begin{array}{c}\text{Intially star have}{10}^{40}\text{deutron}\\ \begin{array}{c}{}_1{H}^2{+}_1{H}^2{⟶}_1{H}^3+\text{Proton}\\ {}_1{H}^2{+}_1{H}^3{⟶}_2{He}^4+\text{Neutron}\end{array}\end{array}$

$\begin{array}{c}\text{and mass}\left({}_1{H}^2\right)=2.014\text{amu}\\ m\left(p\right)=1.007\text{ames}\\ m\left(n\right)=1.008\text{amu}\\ \text{m (}H{e}^4)=4.00\text{amu}\end{array}$

$\begin{array}{c}\text{Power}={10}^{16}\text{watt}\\ \text{Now, our net equation would be}\end{array}$

$3\left({}_1{H}^2\right){⟶}_2^{4}He+n+p$

$\begin{array}{cc}\text{mass defect}\left(\Delta m\right)& =\text{mass of Reactant - mass of product}\\ & =3\left(2.014\right)-(4.00+1.007+1.008)\\ \Delta m& =0.026\text{amu}\end{array}$

$\begin{array}{c}\text{Now, we know}\\ \text{Energy released due to}1\text{amu}=931.5MeV\end{array}$

$\text{Energy released}=931.5\times 0.026\text{Mev}$

$\begin{array}{c}\text{Energy released by}\\ \text{combination of}3\left({{}_{1}H}^2\right)=931.5\times 0.026MeV\end{array}$

$\begin{array}{c}\text{Now,}\\ \begin{array}{cc}\text{Power}& =\frac{\text{Energy}}{t}\\ \text{time}& =\frac{\text{Energy}}{\text{Power}}\end{array}\end{array}$

$\begin{array}{c}=\frac{931.5\times 0.026\times {10}^6\times 1.6\times {10}^{-19}}{{10}^{16}}\\ t=3.9\times {10}^{-28}sec\end{array}$

$\begin{array}{cc}& \text{for}3\text{deutron}\\ & \text{time}=3.9\times {10}^{-28}sec\\ & \text{for}{10}^{40}\text{deutron time}=\frac{3\cdot 9\times {10}^{-28}}{3}\times {10}^{40}\\ & =1.3\times {10}^{12}sec\end{array}$

$t\approx {10}^{12}sec$