Question

A star initially has ${10}^{40}$ deuterons. It produces energy via the processes,

${}_1{\text{H}}^2{+}_1{\text{H}}^2\to {}_1{\text{H}}^3+p$

${}_1{\text{H}}^2{+}_1{\text{H}}^3\to {}_2{\text{He}}^4+n$

The masses of the nuclei are as follows,

$m{(}_1{\text{H}}^2)=2.014\text{amu};m{(}_1{\text{H}}^3)=3.016\text{amu}m\left(p\right)=1.007\text{amu};m\left(n\right)=1.008\text{amu}m{(}_2{\text{He}}^4)=4.001\text{amu}$

If the average power radiated by the star is ${10}^{16}\text{W}$, the deuteron supply of the star is exhausted in a time period of the order of
[Take $1\text{amu}=931.5\frac{\text{MeV}}{c^2}$ ]

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

Answer 1

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

Given,

$m{(}_1{\text{H}}^2)=2.014\text{amu};m{(}_1{\text{H}}^3)=3.016\text{amu}m\left(p\right)=1.007\text{amu};m\left(n\right)=1.008\text{amu}m{(}_2{\text{He}}^4)=4.001\text{amu};P={10}^{16}\text{W}$

${}_1{\text{H}}^2{+}_1{\text{H}}^2\to {}_1{\text{H}}^3+p....\left(1\right)$

${}_1{\text{H}}^2{+}_1{\text{H}}^3\to {}_2{\text{He}}^4+n....\left(2\right)$

Adding (1) and (2) we get,

$3{(}_1{\text{H}}^2){+}_1{\text{H}}^3\to {}_2{\text{He}}^4{+}_1{\text{H}}^3+p+n+Q....(3)$

$\because Q=\Delta m{c}^2$

The mass detect $\left(\Delta m\right)$ is,

$\Delta m=3\times 2.014-4.001-1.007-1.008$

$\therefore \Delta m=0.026\text{amu}$

$\therefore Q=0.026\times 931.5\times {10}^6\times 1.6\times {10}^{-19}$

$Q=3.87\times {10}^{-12}\text{J}$

This is the energy released by $3$ deuterons, then energy released by ${10}^{40}$ deuterons will be,

$=\frac{{10}^{40}\times 3.87\times {10}^{-12}}{3}$

$=1.29\times {10}^{28}\text{J}$

Hence, $\text{Power}=\frac{\text{Energy}}{\text{t}}$

$\Rightarrow t=\frac{\text{Energy}}{\text{Power}}=\frac{1.29\times {10}^{28}}{{10}^{16}}$

$\therefore t=1.29\times {10}^{12}\text{s}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.