| A star initially has \[{{10}^{40}}\]decuterons. It produces energy via the processes |
| \[{}_{1}{{H}^{2}}+{}_{1}{{H}^{2}}\to {}_{1}{{H}^{3}}+P\] |
| \[{}_{1}{{H}^{2}}+{}_{1}{{H}^{3}}\to {}_{1}H{{e}^{4}}+n\] |
| The masses of the nuclei are as follows: |
| \[M({{H}^{2}})=2.014amu;=1.007amu;\] |
| \[M(n)=1.008amu;M(H{{e}^{4}})=4.001amu\] |
| 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 |
A) \[{{10}^{6}}\sec \]
B) \[{{10}^{^{8}}}\sec \]
C) \[{{10}^{^{12}}}\sec \]
D) \[{{10}^{16}}\sec \]
Correct Answer: C
Solution :
| mass defect, \[\Delta M=3\times 2.014-04.001-1.007-1.008\]\[=0.026\,amu\] |
| \[=0.026\times 931\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\]\[=3.82\times {{10}^{-12}}J\] |
| Power of star \[{{10}^{16}}W\] |
| Number of deuterons used =\[\frac{{{10}^{16}}}{\Delta M}=0.26\times {{10}^{28}}\] |
| Time \[=\frac{{{10}^{40}}}{0.26\times {{10}^{28}}}={{10}^{12}}s.\] |
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