Question:

A nucleus with mass number 240 breaks into two 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 13, 2025
  • 216 MeV

  • 0.9 MeV

  • 9.4 MeV

  • 804 MeV

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The Correct Option is A

Solution and Explanation

To solve this problem, we need to calculate the total gain in the binding energy when a nucleus splits into two fragments. Let's walk through the solution step-by-step.

The essential concept here is the calculation of Binding Energy (BE), which is given by the formula:

\(BE = \text{Binding Energy per Nucleon} \times \text{Number of Nucleons}\)

We have the following information: 

  • The mass number of the original nucleus: \(A = 240\)
  • Binding energy per nucleon of unfragmented nucleus: \(7.6 \, \text{MeV}\)
  • The mass number of each fragment: \(120\)
  • Binding energy per nucleon of each fragment: \(8.5 \, \text{MeV}\)

Step 1: Calculate the Binding Energy of the original nucleus.

\(BE_{\text{original}} = 7.6 \, \text{MeV/nucleon} \times 240 = 1824 \, \text{MeV}\)

Step 2: Calculate the Binding Energy of the fragments.

\(BE_{\text{fragments}} = 8.5 \, \text{MeV/nucleon} \times 120 \, (\text{per fragment}) \times 2 \, (\text{for 2 fragments}) = 2040 \, \text{MeV}\)

Step 3: Calculate the total gain in Binding Energy.

The gain is the increase in total binding energy when the nucleus splits into two fragments, calculated as:

\(\Delta BE = BE_{\text{fragments}} - BE_{\text{original}}\)

\(\Delta BE = 2040 \, \text{MeV} - 1824 \, \text{MeV} = 216 \, \text{MeV}\)

This calculation confirms that the total gain in the binding energy is 216 MeV. The correct option is:

\(216 \, \text{MeV}\)

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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.