Question

A star initially has 10${}^{40}$ deuterons. It produces energy via the process
$H^2{+}_1{H}^2{\to }_1{H}^3+p$ and ${}_1{H}^2{\to }_1{H}^3{\to }_{2}H{e}^4+n$. If the average power radiated by the star is 10${}^{16}$ W, the deuteron supply of the star is exhausted in a time of the order of :

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

Answer 1

The correct option is C $10$ ${}^{12}$ $s$

$\begin{array}{c}\text{Given.}\\ \text{Star has initially}{10}^{40}\text{deuterons,}\\ \text{and Energy released via}\end{array}$

$\begin{array}{c}{{}_{1}H}^2+{{}_{1}H}^2⟶{{}_{1}H}^3+P\\ {{}_{1}H}^2+{{}_{1}H}^3⟶{{}_{2}He}^4+n\end{array}$

$\begin{array}{cc}{}_1{H}^2& =\text{deuteron}\\ {}_1{H}^3& =\text{tritium}\\ p& =\text{proton}\\ n& =\text{neutron}\end{array}$

$\text{adding both}$

$3{,}_1{H}^2{\to }_{2}H{e}^4+P+n$

$\text{So, our}\Delta m\text{(mass defect will) be}$

$\begin{array}{cc}\Delta m& =\text{mass of Reactant}-\text{mass of product}\\ & =3\left({\text{mass of}}_1{H}^2\right)\left[\right(\mathrm{mass\; of}He)+m_p+m_n]\end{array}$

$\begin{array}{cc}\text{Let,}m\left({}_1{H}^2\right)& =2.014\\ m_{He}& =4.001\therefore It\text{Should be given}\\ m_p& =1.007\\ m_n& =1.008\end{array}$

$\begin{array}{c}\Delta m=0.026amu\\ \text{Energy released by consumption of}3\left({{}_{1}H}^2\right)=\Delta {mc}^2\\ =0.026\times 931MeV\\ E=24.206MeV\end{array}$

$\begin{array}{c}\text{So,}\\ \text{Energy released by}{10}^{40}\text{deutron =}\end{array}$

$\frac{24.06}{3}\times {10}^{40}Mev$

$=8.068\times {10}^{40}MeV$

$\begin{array}{c}\therefore \mathrm{1ev}=1.6\times {10}^{-19}J\\ =8.068\times {10}^{40}\times 1.6\times {10}^{-19}\times {10}^{-6}J\\ E=12.9088\times {10}^{28}J\end{array}$

$\begin{array}{c}\text{and given,}\\ \text{Average power radiated}={10}^{+16}\\ \text{Time required}=\frac{E}{\text{Power}}\\ =\frac{1.29088\times {10}^{28}}{{10}^{16}}\end{array}$

$\begin{array}{c}\text{Time}=1.29\times {10}^{12}sec\\ \text{time is order of}{10}^{12}sec\end{array}$