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**Nuclei**

... quantities. Therefore, a. Chapter Thirteen. **NUCLEI**. © **NCERT** not to be
republished ... Chadwick was awarded the 1935 Nobel Prize in **Physics** for his
discovery of ... 13.4 MASS-ENERGY AND **NUCLEAR BINDING ENERGY**. 13.4.1
Mass – ... of a **nucleus** and its constituents, ∆M, is called the **mass defect**, and is
given by.

For more information, see **Nuclei**

We express this massive difference in terms of energy by using Einstein's mass energy relationship

Binding energy \Delta E_d = \Delta mc^2

i) ^{56}_{26} Fe

Atomic mass of ^{56}_{26} Fe = 55.934939 amu

number of protons = 26

number of neutrons = 30

Step 1: Finding the mass defect of the given nucleus

Mass of proton m_H = 1.007825 u

Mass of neutron m_n = 1.008665 u

Mass defect of nitrogen \Delta m = m_H + m_n - m

\Delta m=26*(1.007825)+30*(1.008665)-55.934939

\Delta m=0.528741 amu

Converting atomic mass units to MeV/c^2

1 amu = 931.5 MeV/c^2

\Delta m=0.528741*931.5MeV/c^2

\Delta m=492.26\ MeV/c^2

Step 2: Determining binding energy of the given nucleus

Binding energy \Delta E_d = \Delta mc^2

\Delta E_d=492.26MeV/c^2*c^2

\Delta E_d=492.26\ MeV

Therefore, binding energy of the given ^{56}_{26} Fe nucleus \Delta E_d=492.26\ MeV

(ii) ^{209}_{83} Bi

Atomic mass of ^{209}_{83} Bi = 55.934939 amu

number of protons = 83

number of neutrons = 126

Step 1: Finding the mass defect of the given nucleus

\Delta E_d = 104.67 MeV Mass defect of nitrogen \Delta m = m_H + m_n - m

\Delta m=83*(1.007825)+206*(1.008665)-208.980388

\Delta m\ =\ 1.761922\ amu

\Delta m = 1640.26 MeV/c^2

Step 2: Determining binding energy of the given nucleus

\Delta E_d=1640.26\ MeV/c^2*c^2

\Delta E_d=1640.26\ MeV

Hence, binding energy of the given ^{209}_{83} Bi nucleus \Delta E_d=1640.26\ MeV