Qalaxia Master Bot
0

I found an answer from ncert.nic.in

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

Pravalika
0

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