Question:

A nucleus with mass number $240$ breaks into two a fragments each of mass number $120$, the binding energy per nucleon of unfragmented nuclei is $7.6\, MeV$ while that of fragments is $8.5\, MeV$. The total gain in the Binding Energy in the process is :

Updated On: Nov 14, 2025
  • 0.9 MeV
  • 9.4 MeV
  • 804 MeV
  • 216 MeV
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The Correct Option is D

Solution and Explanation

To find the total gain in binding energy after the nuclear reaction, we need to calculate the difference in total binding energy between the fragment nuclei and the original nucleus.

  1. The total binding energy of the original nucleus with mass number $A = 240$ and binding energy per nucleon $7.6 \, \text{MeV}$ is calculated as:
    • Total binding energy of the original nucleus = $7.6 \, \text{MeV/nucleon} \times 240 \, \text{nucleons} = 1824 \, \text{MeV}$
  2. Each of the two fragments has a mass number of $A = 120$, and the binding energy per nucleon is $8.5 \, \text{MeV}$.
    • Total binding energy of one fragment = $8.5 \, \text{MeV/nucleon} \times 120 \, \text{nucleons} = 1020 \, \text{MeV}$
    • Total binding energy of both fragments = $2 \times 1020 \, \text{MeV} = 2040 \, \text{MeV}$
  3. Total gain in binding energy due to the splitting is the difference between the binding energy of the fragments and the binding energy of the original nucleus:
    • Gain in binding energy = Total binding energy of fragments - Total binding energy of original nucleus
    • = $2040 \, \text{MeV} - 1824 \, \text{MeV} = 216 \, \text{MeV}$

Thus, the total gain in the binding energy in the nuclear fission process is 216 MeV. This matches the provided correct answer.

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Concepts Used:

Nuclear Physics

Nuclear physics is the field of physics that studies atomic nuclei and their constituents and interactions, in addition to the study of other forms of nuclear matter. Nuclear physics should not be confused with atomic physics, which studies the atom as a whole, including its electrons

Radius of Nucleus

‘R’ represents the radius of the nucleus. R = RoA1/3

Where,

  • Ro is the proportionality constant
  • A is the mass number of the element

Total Number of Protons and Neutrons in a Nucleus

The mass number (A), also known as the nucleon number, is the total number of neutrons and protons in a nucleus.

A = Z + N

Where, N is the neutron number, A is the mass number, Z is the proton number

Mass Defect

Mass defect is the difference between the sum of masses of the nucleons (neutrons + protons) constituting a nucleus and the rest mass of the nucleus and is given as:

Δm = Zmp + (A - Z) mn - M

Where Z = atomic number, A = mass number, mp = mass of 1 proton, mn = mass of 1 neutron and M = mass of nucleus.