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 1 kg of deuterium undergoes complete fusion, find the amount of total energy released.( 1 amu$=931.5MeV/{{c}^{2}}$)
A. $\approx 6.02\times {{10}^{13}}J$
B. $\approx 5.6\times {{10}^{13}}J$
C. $\approx 9\times {{10}^{13}}J$
D. $\approx 0.9\times {{10}^{13}}J$

Answer 1

Hint: Find the mass defect, to know the actual mass of reactants which will be later used to find the energy released.
Find the Number of nuclei, in 1kg deuterium to calculate the energy released.
Find the Energy released by the formula of energy derived by Einstein.

Complete step-by-step answer:
Mass defect ($\Delta m$ )=mass of reactants – mass of products.
Therefore,$\Delta m$=2$\times$ 2.015-(3.017+1.009),
$\Delta m$= 0.004 amu.
Now,
We know that number of nuclei in 2g deuterium= Avogadro’s number($6.022\times {{10}^{23}}$)
Number of nuclei, in 1kg deuterium is=$\dfrac{6.022\times {{10}^{23}}}{2}\times 1000$ ,
$3.001\times {{10}^{26}}$ nuclei.
Now, according to the problem,
Energy released=$\Delta m{{c}^{2}}$
     \[=0.004\times 931.5\dfrac{MeV}{{{{\not{c}}}^{2}}}\times {{\not{c}}^{2}}\]

=3.726 MeV (this is the energy released by 2g deuterons)
Therefore, energy released by one gram deuteron=$\dfrac{3.726}{2}$ MeV=1.863MeV
Therefore, energy released by 1kg deuteron=$3.001\times {{10}^{26}}\times 1.863$MeV
$=5.6\times {{10}^{26}}MeV$

Now, converting it to joules,
$=5.6\times {{10}^{26}}\times {{10}^{6}}\times 1.6\times {{10}^{-19}}J$
$=8.96\times {{10}^{13}}J\approx 9\times {{10}^{13}}J$
Option C, is the correct answer.

Additional Information:
Nuclear fusion is a reaction in which two or more atomic nuclei are combined to form one or more different atomic nuclei and subatomic particles. The difference in mass between the reactants and products is balanced as either the release or absorption of energy.
One mole of a substance is equal to 6.022 × 10²³ units of that substance, this is known as Avogadro’s number.

Note: Generally at first we have to get the energy released by 2g deuteron is found and then by unitary method we get the mass of 1Kg deuteron, Mass defect formula is necessary to remember.