Calculate the energy (in MeV) released in the following fusion reaction;
`\ _{1}^{2}H` + `\ _{1}^{3}H` → `\ _{2}^{4}He` + `\ _{0}^{1}n`
 (Ans : 17.6 MeV/event)



Given Data:


Mass of Deuterium `\ _{1}^{2}H` = `\m_D` = 2.014102 u

Mass of Tritium `\ _{1}^{3}H` = `\m_T` = 3.01605 u

Mass of Hydrogen `\ _{2}^{4}He` = `\m_{He}` = 4.002603 u

Mass of Oxygen `\ _{0}^{1}n` = `\m_{n}` = 1.008665 u



To Find:

The energy released during the reaction = E = ?



Solution:

We are given

`\ _{7}^{14}N` + `\ _{2}^{4}He` → `\ _{8}^{17}O` + `\ _{1}^{1}H`

and we know that 

△m = (Total Mass of the Reactants ) - (Total Mass of the products )

or

△m = ( `\m_{D}` + `\m_{T}` ) - (`\m_{He}` + `\m_n` )

by putting values

△m = ( 2.014102 u + 3.01605 u) - (4.002603 u + 1.008665 u )

△m = 5.030152 u - 5.011268 u

△m = 0.018884 u


Now 

E = Δm  ｘ 931 MeV

E = 0.018884 u ｘ 931 MeV

E = 17.581004 MeV

E = 17.6 MeV ------------------Ans

Thus the amount of 17.6 MeV of energy is released during the reaction.



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