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JEE Main 2024 (8 Apr, Shift 2) · Mass Defect & Binding Energy

Mass Defect & Binding Energy question from JEE Main 2024 (8 Apr, Shift 2)

Q10JEE Main 2024 · 8 APR · SHIFT 2Easy

Mass Defect & Binding Energy

If M0M_0 is the mass of isotope 512B,Mp{}_{5}^{12}B, M_p and MnM_n are the masses of proton and neutron, then nuclear binding energy of isotope is:

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Answer: (a)

  1. 512B{}_{5}^{12}B has 5 protons and 12−5=712 - 5 = 7 neutrons.
  2. Mass defect Δm=5Mp+7Mn−M0\Delta m = 5M_p + 7M_n - M_0, and BE=Δm c2BE = \Delta m\, c^2
  3. So the answer is (a) (5Mp+7Mn−M0)c2(5M_p + 7M_n - M_0)c^2.
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