Question

The following fusion reaction takes place
$2_1^{2}A{\to }_2^{3}B+n+3.27MeV$
If 2 kg of A is subjected to above reaction, the energy released is used to light a 100 W lamp, how long will the lamp grow ?

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

Answer 1

Answer: (C)

$5\times {10}^4$ years

Number of atoms in 2g of $A=6.023\times {10}^{23}$
$\therefore$ No. of atoms in 2 kg of $A=\frac{6.023\times {10}^{23}}{2}\times 2000$
$=6.023\times {10}^{26}$
$\because$ Energy released in the fusion of 2A nuclei = 3.27 MeV\therefore $Energyreleasedinthefusionof2kgofA$
$E=\frac{327}{2}\times 6.023\times {10}^{23}\times 16\times {10}^{-13}J$
$=15.76\times {10}^{13}J$
$Powerofbulb,P=100W,=100J/s$
$\therefore$ Time for which the bulb will glow $t=\frac{E}{P}$
$=\frac{15.75\times {10}^{13}}{100}$
$\frac{15.75\times {10}^{13}}{100}s$
$=15.75\times {10}^{11}s$=\dfrac{15.75\times 10^{11}}{60 \times 60\times 24 \times 365}$
$=5\times {10}^4$ years