Deuterium undergoes the following fusion reaction . . .

Question : Deuterium undergoes the following fusion reaction : 

H(2,1) + H(2,1) → He(3,2) + n(1,0) + 3.27 MeV

How long an electric bulb of 200 W will glow by using the energy released in 2 g of deuterium? 

Doubt by Krish

Solution : 

H(2,1) + H(2,1) → He(3,2) + n(1,0) + 3.27 MeV

Power of Bulb (P) = 200 W

Mass of Deuterium (m) = 2 g
Molar mass of Deuterium (M) = 2 u

No. of moles (n) 
= Given Mass (m) / Molar Mass (M)
= m/M
= 2/2 
= 1 mole

We know, 1 mole of Deuterium contains, 6.023×10^²³ atoms

Two atoms of Deuterium are used to release energy of 3.27 MeV (Given)
One atom of Deuterium are used to release energy of (3.27/2) MeV

1 mole of atoms of Deuterium are used to release energy of [(3.27/2)×6.023×10²³] MeV

Total energy released in 2g of Deuterium 
= [(3.27/2)×6.023×10²³] MeV
= [(3.27/2)×6.023×10²³] MeV

Also 

Power = Energy / Time
Time = Energy / Power
Time = {[(3.27/2)×6.023×10²³]×10^6×1.6×10^-19}/200
[1M=10⁶, 1eV=1.6×10^-16 J]

= {[(3.27/2)×6.023×1.6]×10¹⁰}/200
= [3.27×6.023×16/4000]×10¹⁰
= [3.27×6.023×4]×10⁷
= 13.08×6.023×10⁷
= 78.78×10⁷
=7.878×10⁸
=7.88×10⁸ Seconds

Similar 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 H(2,1) + H(2,1) → He(3,2)+n+3.27 MeV

Ans : 1.57×10¹² s or 50000 Years (approximately)