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.



************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.