# 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 - Physics

Numerical

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

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

#### 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)/100s

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

Concept: Nuclear Energy - Nuclear Fusion – Energy Generation in Stars
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Chapter 13: Nuclei - Exercise [Page 464]

#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.19 | Page 464
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 19 | Page 464

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