Question

A star has ${10}^{40}$ deuterons. It produces energy via the process
${}_1{H}^2+{}_1{H}^2\to {}_1{H}^3+{}_1{H}^1$
${}_1{H}^2+{}_1{H}^3\to {}_{2}H{e}^4+{}_0{n}^1$
If the average power radiated by the star is ${10}^{16}$ W, then 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 ${10}^{12}s$

The mass defect $\Delta m=1.11\times {10}^{-53}$kg
The energy produced $E=\Delta m{C}^2$
$E=1.11\times {10}^{-53}\times (3\times {10}^8{)}^2={10}^{-36}J$
For ${10}^{40}$ deuterons, the energy produced
$={10}^{40}\times {10}^{-36}J={10}^{4}J$
The average power radiated by star is ${10}^{16}W={10}^{16}Js$
The deuteron supply of the star is exhausted in a time of the order of $\frac{{10}^{16}Js}{{10}^{4}J}={10}^{12}s$