Atoms & Nuclei

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183 of 239

Question 183 of 239

The binding energy per nucleon of 83209Bi^{209}_{83}Bi is _______ MeV.
[Take m(83209Bi)=208.980388 u, mp=1.007825 u, mn=1.008665 u, 1 u=931 MeV/c2][\text{Take } m(^{209}_{83}Bi) = 208.980388 \, u, \, m_p = 1.007825 \, u, \, m_n = 1.008665 \, u, \, 1 \, u = 931 \, \text{MeV}/c^2]

Show answer and solution

Answer: (B) 7.847.84

For the nucleus 83209Bi^{209}_{83}Bi, the number of protons is Z=83Z = 83 and the number of neutrons is N=209−83=126N = 209 - 83 = 126.

Mass of
8383 protons =83×1.007825=83.649475 u= 83 \times 1.007825 = 83.649475 \, u

Mass of
126126 neutrons =126×1.008665=127.091790 u= 126 \times 1.008665 = 127.091790 \, u

Total mass of the nucleons
=83.649475+127.091790=210.741265 u= 83.649475 + 127.091790 = 210.741265 \, u

The mass defect
Δm\Delta m is given by:

Δm=Total mass of nucleons−Mass of nucleus\Delta m = \text{Total mass of nucleons} - \text{Mass of nucleus}

Δm=210.741265−208.980388=1.760877 u\Delta m = 210.741265 - 208.980388 = 1.760877 \, u

The total binding energy (
BEBE) is:

BE=Δm×931 MeVBE = \Delta m \times 931 \, \text{MeV}

BE=1.760877×931=1639.376 MeVBE = 1.760877 \times 931 = 1639.376 \, \text{MeV}

The binding energy per nucleon is:


BEA=1639.376209≈7.84 MeV\dfrac{BE}{A} = \dfrac{1639.376}{209} \approx 7.84 \, \text{MeV}

Answer:
7.847.84

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