Solved Numerical Problem No. 21.5 ( Unit-21:Nuclear Physics); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

Determine the energy associated with the following reaction: (mass of `\ _{6}^{14}C`=14.0077 u)
`\ _{6}^{14}C`  → `\ _{7}^{14}N` + `\ _{-1}^{0}e`
(Ans : 3.77 MeV )

Given Data:

Mass of Carbon `\ _{6}^{14}C` = `\m_C` = 14.0077 u

Mass of Nitrogen `\ _{7}^{14}N` = `\m_{N}` = 14.0031 u

Mass of electron `\ _{-1}^{0}e` = `\m_{e}` = 0.00055 u

To Find:

Energy associated with the given reaction = E = ?

Solution:

We are given

`\ _{6}^{14}C`  → `\ _{7}^{14}N` + `\ _{-1}^{0}e`

and we know that 

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

or

△m =  `\m_{C}`  - (`\m_{N}` + `\m_e` )

by putting values

△m = 14.0077 u - (14.0031 u + 0.00055 u )

△m = 14.0077 u - 14.00365 u

△m = 0.00405 u

Now 

E = Δm  ｘ 931.5 MeV

E = 0.00405 u  ｘ 931 MeV

E = 3.771 MeV ------------------Ans

Thus the amount of 3.771 MeV of energy will be released during the reaction.

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