Question

How long can an electric lamp of 100W be kept glowing by fusion of2.0 kg of deuterium? Take the fusion reaction as2 2 31 1 2 H+ H He+n+3.27 MeV

Answer 1

Given, the power of an electric lamp is 100 W and the mass of deuterium is 2.0 kg.

The fusion reaction is represented as,

H 1 2 + H 1 2 → H 1 2 e+n+3.27 MeV.

1mole of deuterium has a mass of 2 g and contains 6.023× 10 23  atoms. Thus, the number of atoms in 2 kg of deuterium is given by,

n= 6.023× 10 23 2 ×2000 =6.023× 10 26 atoms

When two atoms of deuterium fuse together, then 3.27 MeVenergy is released as shown in the above reaction.

Therefore, total energy per nucleus released in the fusion reaction is

E= 3.27 2 ×6.023× 10 26  MeV …… (1)

Substituting the value of the charge of an electron in the equation (1), we get: E= 3.27 2 ×6.023× 10 26 ×1.6× 10 −19 × 10 6 =1.576× 10 14  J

The total time for which the electric lamp will glow is given by the equation,

t= E P = 1.576× 10 14 100  sec =1.576× 10 12 sec

The total time in years is,

t= 1.576× 10 12 60×60×24×365 =4.9× 10 4

Thus, the total time for which an electric lamp will be kept glowing by the fusion is 4.9× 10 4 years.