Question

The fusion reaction 2H + 2H --> 3He + n releases energy 3.27 MeV per decay. How long could the fusion of 1.0 kg of this hydrogen isotope by this reaction keep a 100 W lamp burning?

Answer 1

time taken is 2.49 x
`10^(5)` year

Step-by-step explanation:

Given data

energy = 3.27 MeV

mass = 1 kg

to find out

time period

solution

we know that deuterium atomic mass unit = 1.66053886 x
`10^(-27)` kg

and we know that energy release = 1 kg × 3.27 / 2×2×1.66053886 x
`10^(-27)`

energy release = 4.9231 x
`10^(26)` MeV

energy release = 4.9231 x
`10^(26)` × 1.6 x
`10^(-13)` = 7.87 x
`10^(14)` J

so at 100 W

energy = power × time

time = 7.87 x
`10^(14)` / 100

time = 7.87 x
`10^(12)` s = 7.87 x
`10^(12)` / 3.15 x
`10^(7)` = 2.49 x
`10^(5)` year

so time taken is 2.49 x
`10^(5)` year