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 $\;10^{40}\;$ deutrons . If produces energy via the process $\;_{1}H^{1}+_{1}H^{2} \to _{1}H^{3}+P\;$ and $\;_{1}H^{2}+_{1}H^{3} \to _{2}He^{4}\;.$ If the average power radiated by star is $\;10^{16}W\;,$ the deutron supply of star is exhausted in a time of order of \[\] (The masses of nuclei are $\;m(H^2)=2.014\;amu\;,m(p)=1.007 amu , \; m(n)=1.0084 amu ,\; m(H)=4.001 amu))$

$(a)\;10^{6} s\qquad(b)\;10^{8} s\qquad(c)\;10^{12} s\qquad(d)\;10^{16} s $

Can you answer this question?

1 Answer

0 votes
Answer : (c) $\;10^{12} s$
Explanation :
$_{1}H^{2}+_{1}H^{2} \to _{1}H^{3}+P$
$_{1}H^{2}+_{1}H^{3} \to _{2}He^{4}+n$
$3\;_{1}H^{2} \to _{2}He^{4}+P+n$
$\bigtriangleup \varepsilon=m\;(_{2}He^{4})m(p)+m(n)-3m\;(_{1}H^{2})$
$\bigtriangleup m=[4.001+1.007+1.008-3(2.014)] amu$
$\bigtriangleup m=-0.026 amu$
$|\bigtriangleup \varepsilon|=C^2\;|\bigtriangleup m|$
$|\bigtriangleup \varepsilon|=(9 \times10^{16})\;(0.026\times1.67\times10^{-27})$
$\bigtriangleup \varepsilon=(931.5)\;(0.026)\;MeV$
$\bigtriangleup \varepsilon=3.87 \times10^{-12}\;J$
As each reaction involves 3 deutrons , so total number of reactions involved in process = $\;\large\frac{10^{40}}{3}\;$if each reaction produces an energy $\;\bigtriangleup \varepsilon\;,$ then
$\varepsilon_{total}=\large\frac{10^{40}}{3}\; \bigtriangleup \varepsilon =1.29\times10^{28} J $
Time of exhaustion of star
$t=1.29\times10^{12} s\;.$
answered Feb 25, 2014 by yamini.v

Related questions

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