Question 613: Atoms & Nuclei

Question

In the fusion reaction,
$$ { }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \longrightarrow{ }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n} $$
the masses of deuteron, helium and neutron expressed in amu are 2.015, 3.017 and 1.009, respectively. If 1 kg of deuterium undergoes complete fusion, then find the amount of total energy released. ( $1 \mathrm{amu}=9315 \mathrm{MeV}$ )

Accepted Answer

Given, fusion reaction,
$$ { }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \longrightarrow{ }_2^3 \mathrm{He}+{ }_0^1 n $$
Mass defect,
$$ \begin{aligned} & \Delta m=\left[2 	imes m\left({ }_1 H^2ight)-\left[m\left({ }_2 \mathrm{He}^4ight)+\mathrm{Mn}ight]ight] \ & =[(2.015 	imes 2)-(3.017+1.009)]=4 	imes 10^{-3} a m u \end{aligned} $$
Equivalent energy, $E=\Delta m c^2$
$$ \begin{aligned} & =4 	imes 10^{-3} 	imes 931.5 \frac{\mathrm{MeV}}{c^2} 	imes c^2 \ & =4 	imes 10^{-3} 	imes 931.5=3.726 \mathrm{MeV} \end{aligned} $$
Number of atoms in 1 kg of ${ }_1^2 \mathrm{H}$
$$ \begin{aligned} & =\frac{	ext { Given mass }}{	ext { Atomic mass }} 	imes 	ext { Avogadro Number } \ & =\frac{1000}{2} 	imes 6.023 	imes 10^{23} \end{aligned} $$
In one reaction, two atoms of ${ }_1^2 \mathrm{H}$ are used.
So, total energy released
$$ \begin{aligned} & =\frac{1}{2}\left(\frac{1000}{2} 	imes 6.023 	imes 10^{23}ight) 	imes 3.726 	imes 1.6 	imes 10^{-13} \ & \simeq 9 	imes 10^{13} \mathrm{~J} \end{aligned} $$

Answer

$9 	imes 10^{13} \mathrm{~J}$

Answer

$20 	imes 10^5 \mathrm{~J}$

Answer

$4 	imes 10^{22} \mathrm{~J}$

Answer

$5 	imes 10^{15} \mathrm{~J}$

Atoms & Nuclei

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