Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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

${ }_1 \mathrm{H}^2$ $+$ ${ }_1 \mathrm{H}^2$ $\rightarrow$ ${ }_1 \mathrm{H}^3$ $+$ $\text { P }$ $\text { and }$

${ }_1 \mathrm{H}^2$ $+$ ${ }_1 \mathrm{H}^3$ $\rightarrow$ ${ }_2 \mathrm{He}^4$ $+$ $\mathrm{n}$

$\begin{aligned} & \mathrm{M}\left({ }_2 \mathrm{He}^4\right)=4.0026034, \mathrm{M}(\text { proton })=1.007825, \mathrm{M}(\text { neutron })=1.008665, \\ & \mathrm{M}(\text { deutron })=2.013553,1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \end{aligned}$

If the average power radiated by the star is $10^{16} \, \, \mathrm{w}$ , the deuteron supply of the star is exhausted in a time of the order of :
Option: 1
$10^6 \mathrm{~s}$

Option: 2
$10^8 \mathrm{~s}$

Option: 3
$10^{12} \mathrm{~s}$

Option: 4
$10^{16} \mathrm{~s}$

Answers (1)

best_answer

$3 \ { }_{1} \mathrm{H}^2 \rightarrow{ }_2 \mathrm{He}^4+\mathrm{P}+\mathrm{n} \quad \Delta \mathrm{m}=0.0216$

$\Rightarrow \mathrm{E}=(\Delta \mathrm{m}) \times 931.4 \mathrm{MeV}$

$=3.2 \times 10^{-12} \mathrm{~J}$

$\Rightarrow \mathrm{E}_{\text {total }}=\frac{\mathrm{E} \times 10^{40}}{3} \approx 10^{28} \quad \Rightarrow \mathrm{t} \approx \frac{10^{28}}{10^{16}} \approx 10^{12}$

Posted by

avinash.dongre

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

‘Re-NEET must be secure, transparent and fool proof’: Dharmendra Pradhan at review meeting

anam-khan

NEET UG 2026 Re-exam on June 21; top medical colleges with lowest MBBS fees

anam-khan

Re-NEET Exam 2026 LIVE: Re-test on June 21; CBI probes NTA leak link, official guidelines, correction window

Latest Question