Solved Numerical Problem No. 21.4 ( Unit-21:Nuclear Physics); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

Find the energy associated with the following reaction: (M ass of `\ _{1}^{1}H`=1.00784 u)

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

What does a negative sign indicate? (Ans: 1.12 MeV)

Given Data:

Mass of Hydrogen `\ _{1}^{1}H` = `\m_H` = 1.00784 u

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

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

Mass of Oxygen `\ _{8}^{17}O` = `\m_{O}` = 16.9991 u

To Find:

Energy associated with the given 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_{N}` + `\m_{He}` ) - (`\m_{O}` + `\m_H` )

by putting values

△m = ( 14.0031 u + 4.002603 u) - (16.9991 u + 1.00784 u )

△m = 18.005703 u - 18.00694 u

△m = -0.001237 u

0r

△m = -1.237 ｘ10⁻³ u

Now 

E = Δm  ｘ 931.5 MeV

E = -1.2 ｘ10⁻³ u  ｘ 931.5 MeV

E = -1.12 MeV ------------------Ans

The negative shows that the amount of 1.12 MeV of energy is required to start the reaction.

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