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QJEEM24Ph2f03: JEE Main 2024 Physics - Class 12 JEEM24Ph2f03 Solution: If is the mass of isotope and are the masses of proton and…

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:

Multiple Choice Options

  • A (5Mp+7Mn−Mo)C2(5 M_p+7 M_n-M_o) C^2
  • B (Mo−5Mp−7Mn)C2(M_o-5 M_p-7 M_n) C^2
  • C (Mo−5Mp)C2(M_o-5 M_p) C^2
  • D (M0−12Mn)C2(M_0-12 M_n) C^2

Step-by-Step Verified Solution

To determine the nuclear binding energy of the isotope 512B_5^{12}B , we need to consider the mass defect concept. The mass defect is the difference between the sum of the individual masses of nucleons (protons and neutrons) and the actual mass of the nucleus.

Let's calculate the mass defect first. The isotope 512B_5^{12}B has 5 protons and 7 neutrons (since the total number of nucleons is 12). Therefore, the mass defect can be written as:

ΔM=(5Mp+7Mn)−M0\Delta M = (5 M_p + 7 M_n) - M_0

Once we have the mass defect, the binding energy can be found using Einstein's mass-energy equivalence principle, which is given by:

E=ΔM⋅c2E = \Delta M \cdot c^2

Substituting the mass defect into this equation, we get:

E=((5Mp+7Mn)−M0)⋅c2E = ((5 M_p + 7 M_n) - M_0) \cdot c^2

Therefore, the correct answer is Option A:

(5Mp+7Mn−M0)⋅c2(5 M_p + 7 M_n - M_0) \cdot c^2

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