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

Answer:

time taken is  2.49  x $10^{5}$ year

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