Question

Deuterium undergoes the following fusion reaction:

\[\ce{_1^2H + _1^2H -> _2^3He + _0^1n + 3.27 MeV}\]

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

Answer 1

Deuterium \[\ce{_1H^2}\]

Given mass = 2g

Atomic weight = 2

Number of atom in 2 g of deuterium 

= `(6.023 xx 10^23 xx 2)/2`

= 6.023 × 10²³

Energy released in 2 g Deuterium = 3.27 MeV

Total energy = `3.27/2 xx 6.023 xx 10^23` MeV

= `3.27/2 xx 6.023 xx 10^23 xx 10^6 xx 1.6 xx 10^-19` J

= 15.75 × 10¹⁰ J

Power consumed by bulb = 200 W

Time for bulb glow = `(15.75 xx 10^10)/200`

= 7.87 × 10⁸ s

= `(7.87 xx 10^8)/(365 xx 24 xx 60 xx 60)` year

= 24.95 year