Solved Numerical Problem No.2 ( Unit-20:Nuclear Physics); HSSC-II (Class-12); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

The mass of the nucleus is 13.999234u, calculate the binding energy. (Answer: 104.66 MeV)  

Given Data:

Mass of the Nucleus = m = 13.999234 u Shows it is a Nitrogen (N-14) Atom. So,


Atomic mass of helium = A = 14

Atomic number of helium = z = 7

Mass of Proton = `\m_p` = 1.00728 u

Mass of Neutron = `\m_n`= 1.00867 u

To Find:

Binding energy = B.E =?

Solution:

The formula for binding energy is

B.E. = Δmc², 

Where Δm should be in kg unit so, to find the mass defect Δm we have

 Δm = ( Z `\m_p` + (A - Z)`\m_n` ) - m

by putting the corresponding values

 Δ m = ( 7 ｘ 1.00728 u + (14 - 7) ｘ1.00867 u  - 13.999234 u

 Δ m = 7.05096 u + 7.06069 u - 13.999234 u

Δ m = 0.112416 u

Now To Find the Binding energy = B.E =?

We know that  1 amu = 931.5 MeV so,

Binding Energy = B.E = Mass deficit (in amu) ×  931.5 MeV

Binding Energy = B.E = 0.112416 u ×  931.5 MeV

B.E = 104.715504 MeV -----------------------Ans.2

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