Question

A star initially has ${10}^{40}$ deuterons. It produces energy via the processes
${}_1{\text{H}}^2{+}_1{\text{H}}^2{\to }_1{\text{H}}^3+p\text{and}{}_1{\text{H}}^2{+}_1{\text{H}}^3{\to }_2{\text{He}}^4+n$. If the average power radiated by the star is ${10}^{16}\text{W}$, the deuteron supply of the star is exhausted in a time of the order of $11.6\times {10}^x\text{s}$. Then the value of $x$ is -

$\text{Take}m{(}_1{\text{H}}^2)=2.0141u,m{(}_2{\text{He}}^4)=4.0026u,m{(}_1{p}^1)=1.0078u,m{(}_0{n}^1)=1.0086u1u{c}^2=931.5\text{MeV}$

• A. $7$
• B. $9$
• C. $11$
• D. $13$

Answer 1

The correct option is C $11$


Mass defect :
$\Delta m=[3\times 2.0141u-(4.0026u+1.0078u+1.0086u\left)\right]$

$\Rightarrow \Delta m=0.0233u$

$\Rightarrow \text{Equivalent energy}=\Delta m{c}^2=0.0233\times 931.5\text{MeV}$

$=21.7\text{MeV}=21.7\times 1.6\times {10}^{-13}\text{J}$

$=34.7\times {10}^{-13}\text{J}$

${\text{Number of}}_1{\text{H}}^2\text{atoms involved in reactions}=3$

Therefore, number of reactions possible $=\frac{{10}^{40}}{3}$

$\Rightarrow \text{Total energy released}=\frac{{10}^{40}}{3}\times 34.7\times {10}^{-13}=11.6\times {10}^{27}\text{J}$

We know that, $P=\frac{E}{t}$

$\Rightarrow t=\frac{E}{P}=\frac{11.6\times {10}^{27}}{{10}^{16}}$

$\Rightarrow t=11.6\times {10}^{11}$

Given that $t=11.6\times {10}^x\text{s}$

$\therefore x=11$