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:
${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{3}{He} + n + 3.2 MeV .$

Solution

Power of the electric lamp = 100 W

When two nuclei of deuterium fuse together, energy released = 3.2 MeV

Number of deuterium atoms in 2 kg is,
$= \frac{6.023 \times 10^{23}}{2} \times 2000 = 6.023 \times 10^{26}$

Energy released when $6.023 \times 10^{26}$ nuclei of deuterium fuse together,

$= \frac{3.2}{2} \times 6.023 \times 10^{26} MeV = \frac{3.2}{2} \times 6.023 \times 10^{26} \times 1.6 \times 10^{- 13} J = 1.54 \times 10^{14} J or Ws$
If the lamp glows for time t, then the electrical energy consumed by the lamp is 100 t,

$∴$ 100 t = $1.54 \times 10^{14}$

$t = 1.54 \times 10^{12} s$

$= \frac{1.54 \times 10^{12}}{3.154 \times 10^{7}} years = 4.88 \times 10^{4} years .$

which is the life span of an electric lamp.

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Wave Optics

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:
${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{3}{He} + n + 3.2 MeV .$

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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:H12 + H12 → He23+n+3.2 MeV.

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Mock Test Series

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Entrance Exams Preparation

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:
${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{3}{He} + n + 3.2 MeV .$