How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as :
$_{1}^{2} H +_{1}^{2} \to_{2}^{3} H e + n + 3.27 M e V$ .

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Solution

Number of atoms in 2 kg deuterium = $\frac{6.023 \times 10^{23} \times 2000}{2} = 6.023 \times 10^{26}$
Energy released when 2 atoms fuse = $3.27 M e V$
Thus, total energy released = $\frac{3.27}{2} \times 6.023 \times 10^{26} M e V$ $= 15.75 \times 10^{13} J$ .
Energy consumed by bulb per second= $100 J$ .
Thus, time for which the bulb glows = $\frac{15.75 \times 10^{13}}{100} s$ $= 15.75 \times 10^{11} s = 4.99 \times 10^{7} y e a r s$ .

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How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

2_{1}^{2} A \to_{2}^{3} B + n + 3.27 M e V

1

H_{2} O

_{1}^{2} H

_{1}^{2} H

\to

_{1}^{3} H

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1.5 \times 10^{2}

_{1}^{2} H +_{1}^{2} H \to_{2}^{3} H e +_{0}^{1} n

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