Question

The following fusion reaction takes place
$ {}_{1}^{2} H +{}_{1}^{2}H \to {}_{2}^{3} He + n +3.27 \,MeV $
If $ 2\, kg $ of deuterium is subjected to above reaction, the energy released is used to light a $ 100\, W $ lamp. How long will the lamp glow?

• A. $ 2\times10^6 $ year
• B. $ 3\times10^5 $ year
• C. $ 5\times10^4 $ year
• D. $ 7\times10^3 $ year

Answer 1

Answer: (C)

Let $t$ be the time.
According to the Avogadro number concept
Number of atoms in $2g$ of deuterium
$= 6 .0 2 \times 10^{23}$
Number of atoms in $2 \,kg$ of deuterium
$ = \frac{6.023 \times 10^{23} \times 2 \times 10^3}{2}$
$ = 6.023 \times 10^{26}$ Nuclei
From given equation energy released during
fusion of two deuterium $= 3.27 \,MeV$
$\therefore $ Energy released by one deuterium
$ = \frac{3.27}{2} = 1.635 \,MeV$
Energy released in $6.023 \times 10^{26}$ deuterium atoms
$= 1.635 \times 6.023 \times 10^{26}$
$= 9.846 \times 10^{26}\, MeV$
$ = 9.848 \times 10^{26} \times 1.6 \times 10^{-13}$
$ = 15.75 \times 10^{13} \,s$
Time $ = \frac{15.75 \times 1\times 10^{13}}{100}$
$ = 15.75 \times 10^{11} s$
$ = \frac{15.75 \times 10^{11}}{60\times 24 \times 60\times 365}$
$ = 4.97 \times 10^4$
$ = 5\times 10^4 $ years