Modern Physics Ques 188
- A star initially has $10^{40}$ deuterons. It produces energy via the processes ${ } _1 H^{2}+{ } _1 H^{2} \rightarrow{ } _1 H^{3}+p$ and ${ } _1 H^{2}+{ } _1 H^{3} \rightarrow{ } _2 He^{4}+n$. 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
(1993, 2M)
(a) $10^{6} s$
(c) $10^{12} s$
(b) $10^{8} s$
(d) $10^{16} s$
The mass of the nuclei are as follows
$$ M\left(H^{2}\right)=2.014 amu ; M(n)=1.008 amu $$
$$ M(p)=1.007 amu ; M\left(He^{4}\right)=4.001 amu $$
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Answer:
Correct Answer: 188.(c)
Solution:
Formula:
- The given reactions are :
$$ \begin{aligned} { } _1 H^{2}+{ } _1 H^{2} & \rightarrow{ } _1 H^{3}+p \\ { } _1 H^{2}+{ } _1 H^{3} & \rightarrow{ } _2 He^{4}+n \\ 3{ } _1 H^{2} & \rightarrow{ } _2 He^{4}+n+p \end{aligned} $$
Mass defect
$$ \begin{aligned} \Delta m & =(3 \times 2.014-4.001-1.007-1.008) amu \\ & =0.026 amu \end{aligned} $$
Energy released $=0.026 \times 931 MeV$
$$ =0.026 \times 931 \times 1.6 \times 10^{-13} J=3.87 \times 10^{-12} J $$
This is the energy produced by the consumption of three deuteron atoms.
$\therefore$ Total energy released by $10^{40}$ deuterons
$$ =\frac{10^{40}}{3} \times 3.87 \times 10^{-12} J=1.29 \times 10^{28} J $$
The average power radiated is $P=10^{16} W$ or $10^{16} J / s$.
Therefore, total time to exhaust all deuterons of the star will be $t=\frac{1.29 \times 10^{28}}{10^{16}}=1.29 \times 10^{12} s \approx 10^{12} s$
BETA
