Question

A star converts all its hydrogen to helium, achieving a $100\%$ helium composition. Next it converts the helium to carbon via the triple-alpha process,
${}^{4}He{+}^{4}He{+}^{4}He{\to }^{12}C+7.27Mev$
The mass of the star is $4.6\times {10}^{32}kg$, and it generates energy at the rate of $5.3\times {10}^{30}W$.How long will it take to convert all the helium to carbon at this rate?

Answer 1

Fusion takes place deep within the sun is given by

$4\frac{1}{1}H\to \frac{4}{2}He+2e^{+}+2v+26.7MeV$

If $Q_1$ is the energy released by fusion of 1.0 kg of hydrogen, then

$Q_1=\frac{1000\times 6.023\times {10}^{23}\times 26.7}{4}MeV$
$=40.2\times {10}^{26}MeV$

Again,during the fission of one nucleus of ${}^{235}U$, energy released is 200 MeV.
If $Q_2$ be the energy released by the fission of ${10}^3$ of ${}^{235}U$, then

$Q_2=\frac{1000\times 6.023\times {10}^{23}\times 200}{235}MeV$
$=5.1\times {10}^{26}MeV$

$\frac{Q_1}{Q_2}=\frac{40.2\times {10}^{26}}{5.1\times {10}^{26}}=7.88=8$

SO, energy released per kg of fuel in fusion is nearly 8 times that in fission.