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  • Obtain the binding energy of the nuclei 56 26 Fe and 209 83 Bi in units of MeV from the following data: m (83 209 Bi)=208.980388 u

Q. 13.4 (ii) Obtain the binding energy of the nuclei _{26}^{56}\textrm{Fe} and _{83}^{209}\textrm{Bi} in units of MeV from the following data:

                 (ii)m(_{83}^{209}\textrm{Bi})=208.980388\; \; u

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mH = 1.007825 u

mn = 1.008665 u

Atomic mass of _{83}^{209}\textrm{Bi} is m=208.980388 u

Mass defect

\Delta m=(209-83)\timesmn+83\times m_H - m

\Deltam=126\times1.008665+83\times1.007825 - 208.980388

\Deltam=1.760877 u

Now 1u is equivalent to 931.5 MeV

Eb=1.760877 \times931.5

Eb=1640.2569255 MeV

Therefore binding energy of a _{83}^{209}\textrm{Bi} nucleus is 1640.2569255 MeV.

Average\ binding\ energy=\frac{1640.25}{208.98}=7.84MeV

 

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