Question

In the fusion reaction ${}_1^{2}He{+}_1^{2}H{\to }_2^{3}He{+}_0^{1}n$, the masses of deuteron, helium, and neutron expressed in amu are $2.015,3.017$ and $1.009$, respectively. If $1kg$ of deuterium undergoes complete fusion, find the amount of total energy released. $(1amu=931.5meV/c^2)$

• A. $\approx 6.02\times {10}^{13}J$
• B. $\approx 5.6\times {10}^{13}J$
• C. $\approx 9.0\times {10}^{13}J$
• D. $\approx 0.9\times {10}^{13}J$

Answer 1

The correct option is C $\approx 9.0\times {10}^{13}J$

$E=\Delta m{c}^2$
For one reaction:
Mass Defect $=\Delta m$
$=2\left(m_H\right)-m_{He}-m_n$
$=2\left(2.015\right)-3.017-1.009$
$=0.004amu$
$1amu=931.5MeV/c^2$
Hence,
$E=0.004\times 931.5MeV=3.724MeV$
$E=3.726\times 1.6\times {10}^{-13}J=5.96\times {10}^{-13}J$

For $1kg$ of Deuterium available,
$moles=\frac{1000g}{2g}=500$
$N=500N_A=3.01\times {10}^{26}$
Energy released $=\frac{N}{2}\times 5.95\times {10}^{-13}J$
$=8.95\times {10}^{13}J$