Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A nucleus with mass number 240 breaks into 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:

804 MeV

216 MeV

0.9 MeV

9.4 MeV

Step-by-Step Solution

Concept: The total binding energy ( $BE$ ) of a nucleus is the product of its mass number ( $A$ ) and its binding energy per nucleon ( $BE/A$ ). Energy is released (gain in binding energy) when a heavy nucleus splits into lighter, more stable nuclei (fission) with higher binding energy per nucleon .
Calculate Initial Binding Energy ( $BE_i$ ):

Mass number ( $A$ ) = 240
$BE/A$ = 7.6 MeV
$BE_i = 240 \times 7.6 \text{ MeV} = 1824 \text{ MeV}$ .

Calculate Final Binding Energy ( $BE_f$ ):

The nucleus breaks into fragments of mass number 120. Since $240/120 = 2$ , there are two fragments.
Total nucleons remain conserved (240).
$BE/A$ (fragments) = 8.5 MeV
$BE_f = 240 \times 8.5 \text{ MeV} = 2040 \text{ MeV}$ .

Calculate Total Gain in Binding Energy:

Gain = $BE_f - BE_i$
Gain = $2040 \text{ MeV} - 1824 \text{ MeV} = 216 \text{ MeV}$ .
Alternatively: $Gain = A \times (BE/A_{final} - BE/A_{initial}) = 240 \times (8.5 - 7.6) = 240 \times 0.9 = 216 \text{ MeV}$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started