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

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV

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13.19 The given fusion reaction is

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV

Amount of deuterium, m = 2 kg

1 mole, i.e. 2 g of deuterium contains 6.023 $\times 10^{23}$
atoms

Hence 2 kg of deuterium contains =
$\frac{6.023 \times 10^{23}}{2}$ $\times 2 \times 10^{3}$ atoms = 6.023 $\times 10^{26}$ atoms

It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27 MeV of energy is released.

Hence total energy released from 2 kg of deuterium, E =
$\frac{3.27}{2}$ $\times$ 6.023 $\times 10^{26}$ MeV

= $\frac{3.27}{2}$ $\times$ 6.023 $\times 10^{26} \times 10^{6} \times 1.6 \times 10^{- 19}$ J = 1.5756 $\times 10^{14}$ J

Power of the electric bulb, P = 100 W = 100 J/s

Hence energy consumed by the bulb per second = 100 J

Therefore, total time the electric bulb will glow = $\frac{1.5756 \times 10^{14}}{100}$ seconds = 1.5756 $\times 10^{12}$ secs

= 49.96 $\times 10^{3}$ years

...more

13.19 The given fusion reaction is

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV

Amount of deuterium, m = 2 kg

1 mole, i.e. 2 g of deuterium contains 6.023 $\times 10^{23}$
atoms

Hence 2 kg of deuterium contains =
$\frac{6.023 \times 10^{23}}{2}$ $\times 2 \times 10^{3}$ atoms = 6.023 $\times 10^{26}$ atoms

It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27 MeV of energy is released.

Hence total energy released from 2 kg of deuterium, E =
$\frac{3.27}{2}$ $\times$ 6.023 $\times 10^{26}$ MeV

= $\frac{3.27}{2}$ $\times$ 6.023 $\times 10^{26} \times 10^{6} \times 1.6 \times 10^{- 19}$ J = 1.5756 $\times 10^{14}$ J

Power of the electric bulb, P = 100 W = 100 J/s

Hence energy consumed by the bulb per second = 100 J

Therefore, total time the electric bulb will glow = $\frac{1.5756 \times 10^{14}}{100}$ seconds = 1.5756 $\times 10^{12}$ secs

= 49.96 $\times 10^{3}$ years

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

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV

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13.19 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as H12 + H12 → H e 2 3 + n+3.27 MeV

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

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV