Q69 / 75 · JEE Main 2026 (02 April Shift 2) · Physics
The binding energy per nucleon of $^{209}_{83}Bi$ is _______ MeV.
$[\text{Take } m(^{209}_{83}Bi) = 208.980388 \, u, \, m_p = 1.007825 \, u, \, m_n = 1.008665 \, u, \, 1 \, u = 931 \, \text{MeV}/c^2]$
$[\text{Take } m(^{209}_{83}Bi) = 208.980388 \, u, \, m_p = 1.007825 \, u, \, m_n = 1.008665 \, u, \, 1 \, u = 931 \, \text{MeV}/c^2]$
- A$7.48$
- Correct answerB$7.84$
- C$8.79$
- D$6.94$
Solution
For the nucleus $^{209}_{83}Bi$, the number of protons is $Z = 83$ and the number of neutrons is $N = 209 - 83 = 126$.
Mass of $83$ protons $= 83 \times 1.007825 = 83.649475 \, u$
Mass of $126$ neutrons $= 126 \times 1.008665 = 127.091790 \, u$
Total mass of the nucleons $= 83.649475 + 127.091790 = 210.741265 \, u$
The mass defect $\Delta m$ is given by:
$\Delta m = \text{Total mass of nucleons} - \text{Mass of nucleus}$
$\Delta m = 210.741265 - 208.980388 = 1.760877 \, u$
The total binding energy ($BE$) is:
$BE = \Delta m \times 931 \, \text{MeV}$
$BE = 1.760877 \times 931 = 1639.376 \, \text{MeV}$
The binding energy per nucleon is:
$\dfrac{BE}{A} = \dfrac{1639.376}{209} \approx 7.84 \, \text{MeV}$
Answer: $7.84$
Mass of $83$ protons $= 83 \times 1.007825 = 83.649475 \, u$
Mass of $126$ neutrons $= 126 \times 1.008665 = 127.091790 \, u$
Total mass of the nucleons $= 83.649475 + 127.091790 = 210.741265 \, u$
The mass defect $\Delta m$ is given by:
$\Delta m = \text{Total mass of nucleons} - \text{Mass of nucleus}$
$\Delta m = 210.741265 - 208.980388 = 1.760877 \, u$
The total binding energy ($BE$) is:
$BE = \Delta m \times 931 \, \text{MeV}$
$BE = 1.760877 \times 931 = 1639.376 \, \text{MeV}$
The binding energy per nucleon is:
$\dfrac{BE}{A} = \dfrac{1639.376}{209} \approx 7.84 \, \text{MeV}$
Answer: $7.84$
Hence, the correct answer is option: 'B'.