Modern Physics 5 Question 15

15. 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: 15. (c)

Solution:

  1. 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 $t=\frac{1.29 \times 10^{28}}{10^{16}}=1.29 \times 10^{12} s \approx 10^{12} s$



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