Physics Ncert Solutions Class 12th Nuclei Class 12th Physics

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

Option 1 - <p>0.9MeV<br><br></p>

Option 2 - <p>9.4MeV<br><br></p>

Option 3 - <p>804MeV</p>

Option 4 - <p>216MeV</p>

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Answered by

Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 4

Detailed Solution:

The reaction is X²? → Y¹²? + Z¹²?
Binding energies per nucleon are: X=7.6 MeV, Y=8.5 MeV, Z=8.5 MeV.
Gain in binding energy (Q) = (Binding energy of products) - (Binding energy of reactants)
Q = (120 * 8.5 + 120 * 8.5) - (240 * 7.6) MeV
Q = (2 * 120 * 8.5) - (240 * 7.6) MeV = 2040 - 1824 = 216 MeV.

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Physics Ncert Solutions Class 12th 2023

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