Solved Numerical Problem No.6 ( Unit-20:Nuclear Physics); HSSC-II (Class-12); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

Find the energy released when B-decay changes `\ ^{234}Th` into `\ ^{234}Pa`. Mass of `\ ^{234}Th` = 234.0436 u and `\ ^{234}Pa` = 234.0428 u. (Answer: 0.23275 MeV)

Given Data:

Mass of Thorium `\ _{90}^{234}Th` = `\m_Th` = 234.0436 u

Mass of Protactinium `\ _{91}^{234}Pa` = `\m_{Pa}` = 234.0428 u

Mass of  `\ _{-1}^{0}β` = `\m_β` = 0.00055 u

To Find:

The energy released  = E =?

Solution:

According to the conditions, the reaction is

`\ _{90}^{234}Th`   →  `\ _{91}^{234}Pa` + `\ _{-1}^{0}β` + Energy

Before calculating the energy released E in the reaction, we will find the mass Δm of the energy released. 

So from the reaction 

`\m_{Th}` = `\m_{Pa}`  + `\m_β` + Δm

or

Δm   = `\m_{Th}` - (`\m_{Pa}`  + `\m_β`)

by putting values

Δm   = 234.0436 u - (234.0428 u + 0.00055 u) 

Δm   = 234.0436 u - 234.04335 u

Δm  = 0.00025 u 

Now 

E = Δm  ｘ 931 MeV

E = 0.00025 u   ｘ 931 MeV

E = 0.23275 MeV ------------------Ans.

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