118k views
0 votes
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?

User Jarvis
by
7.3k points

1 Answer

4 votes

Answer:

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

User Abdul Ahmad
by
6.7k points