Physics - Modern Physics Question with Solution | TestHub

Number of atoms in$2 \mathrm{g}$of${}_1{}^{2}A=6.023\times {10}^{23}$

$\therefore$No of atoms in$2 \mathrm{kg}$of${}_1{}^{2}A=\frac{6.023\times {10}^{23}}{2}\times 2000$

$\because$Energy released in the fusion of two${}_1{}^{2}A$nuclei$=3.27\mathrm{MeV}$

$\therefore$Energy released in the fusion of$2 \mathrm{kg}$of${}_1{}^{2}A$

$\therefore E=\frac{327}{2}\times 6.023\times {10}^{26}\times 16\times {10}^{-13}\mathrm{J}$

$=15.76\times {10}^{13}\mathrm{J}$

Power of bulb,$P=100\mathrm{W},=100$$\mathrm{J} {\mathrm{s}}^{-1}$

$\therefore$Time for which the bulb will glow$t=\frac{E}{P}$

$=\frac{15.75\times {10}^{13}}{100}$

$=15.75\times {10}^{11}\mathrm{s}$

$=\frac{15.75\times {10}^{11}}{60\times 60\times 24\times 365}$

$=5\times {10}^4$years