Nuclear Physics And Radioactivity Question 122
Question: A star initially has 1040 deuterons. It produces energy via the processes $ _{1}H^{2}+ _{1}H^{2}\to _{1}H^{3}+p $ $ _{1}H^{2}+ _{1}H^{3}\to _{2}H^{4}+n $ The masses of the nuclei are as follows: $ M(H^{2}) $ = 2.014 amu; M (p) = 1.007 amu; $ M(n) $ = 1.008 amu; $ M(He^{4}) $ = 4.001 amu If the average power radiated by the star is $ 10^{16} $ W, the deuteron supply of the star is exhausted in a time of the order of
Options:
A) $ 10^{6} $
B) $ 10^{8}\sec $
C) $ 10^{12}\sec $
D) $ 10^{16}\sec $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Mass defect $ =3\times 2.014-4.001-1.007-1.008 $ $ =0.026amu=0.026\times 931\times 10^{6}\times 1.6\times {{10}^{-19}}J $ $ =3.82\times {{10}^{-12}}J $ Power of star $ =10^{16}W $ Number of deuterons used $ =\frac{10^{16}}{\Delta M}=0.26\times 10^{28} $ Deuteron supply exhausts in $ \frac{10^{40}}{0.26\times 10^{28}}=10^{12}s $
BETA
