Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
0 votes

A star initially has $\;10^{4}\;$ deuterons . It produces energy via the process \[\] $\;_{1}H^{2}+_{1}H^{2} \to _{1}H^{3}+P\;$ and \[\] $\;_{1}H^{2}+_{1}H^{3} \to _{2}He^{4}+n\;$\[\] If average power radiated by star is $\;10^{16} W\;$ the deutron supply of star is exhausted in a time of order of

$(a)\;10^{11} s\qquad(b)\;10^{12} s\qquad(c)\;10^{13} s\qquad(d)\;10^{14} s$

Can you answer this question?

1 Answer

0 votes
Answer : (b) $\;10^{14} s$
Explanation :
The given reactions are
$\;_{1}H^{2}+_{1}H^{2} \to _{1}H^{3}+P\;$
$\;_{1}H^{2}+_{1}H^{3} \to _{2}He^{4}+n\;$
$\; 3 _{1}H^{2} \to _{2}He^{4}+n+P\;$
Mass defect
$\bigtriangleup m=(3\times2.014-4.001-1.007-1,008)amu$
$=0.026 amu$
Energy released $\;=0.026\times931 MeV$
$=0.026\times931\times1.6\times10^{-13} J$
$=3.87 \times10^{-12} J$
This energy produced by consumption of three deuteron atoms .
Total energy released by $\;10^{40}\;$ deuterons .
$=\large\frac{10^{40}}{3}\times3.87\times10^{-12} J$
$=1.29\times10^{28} J$
The average power radiated is $\;P=10^{16} W$
or $\;10^{16} J/s$
Total time to exhaust all deuterons of stars will be
$t=\large\frac{1.29 \times10^{28}}{10^{16}}=1.29 \times 10^{12} s$
$\approx 10^{12} s\;.$
answered Feb 27, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App