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Karnataka Board PUCPUC Science 2nd PUC Class 12

How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as X12X2122H+X12X2122H⟶X13X2123He+n+3.27MeV

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Question

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

\[\ce{^2_1H + ^2_1H -> ^3_1He + n + 3.27 MeV}\]

Numerical
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Solution

The given fusion reaction is:

\[\ce{^2_1H + ^2_1H -> ^3_1He + n + 3.27 MeV}\]

Amount of deuterium, m = 2 kg

1 mole, i.e., 2 g of deuterium contains 6.023 × 1023 atoms.

∴2.0 kg of deuterium contains = `(6.023 xx 10^23)/2 xx 2000 = 6.023 xx 10^26` atoms

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

∴ Total energy per nucleus released in the fusion reaction:

`"E" = 3.27/2 xx 6.023  xx 10^26 "MeV"`

`= 3.27/2 xx  6.023 xx 10^26 xx 1.6 xx 10^(-19) xx 10^6`

`= 1.576 xx 10^14 " J"`

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

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is calculated as:

`(1.576  xx 10^14)/100`s

`(1.576 xx 10^14)/(100 xx 60 xx 60 xx 24 xx 365) ~~ 4.9 xx 10^4` year

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Chapter 13: Nuclei - Exercise [Page 464]

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NCERT Physics Part I and II [English] Class 12
Chapter 13 Nuclei
Exercise | Q 19 | Page 464
NCERT Physics Part I and II [English] Class 12
Chapter 13 Nuclei
Exercise | Q 13.19 | Page 464

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