Question

The following fusion reaction takes place
${}_1^{2}H+{}_1^{2}H\to {}_2^{3}He+n+3.27MeV$
If $2kg$ of deuterium is subjected to above reaction, the energy released is used to light a $100W$ lamp. How long will the lamp glow?

• A. $2\times 10^6$ year
• B. $3\times 10^5$ year
• C. $5\times 10^4$ year
• D. $7\times 10^3$ year

Answer 1

Answer: (C)

Let $t$ be the time.
According to the Avogadro number concept
Number of atoms in $2g$ of deuterium
$=6.02\times 10^{23}$
Number of atoms in $2kg$ 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 $=3.27MeV$
$\therefore$ Energy released by one deuterium
$=\frac{3.27}{2}=1.635MeV$
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