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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:
  • a)
    0.9MeV
  • b)
    9.4MeV
  • c)
    804MeV
  • d)
    216MeV
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A nucleus with mass number 240 breaks into two fragments each of mass ...

BE gain = (120 × 8.5) + (120 × 8.5) − (240 × 7.6)
= 2040 − 1824
= 216 MeV
Free Test
Community Answer
A nucleus with mass number 240 breaks into two fragments each of mass ...
Given:
- Mass number of unfragmented nucleus (A) = 240
- Mass number of each fragment (A1 = A2) = 120
- Binding energy per nucleon of unfragmented nuclei (BE/A) = 7.6 MeV
- Binding energy per nucleon of fragments (BE1/A1 = BE2/A2) = 8.5 MeV

To find:
Total gain in the binding energy in the process

Solution:

Step 1: Calculation of binding energy of unfragmented nucleus:
The total binding energy (BE) of an unfragmented nucleus can be calculated using the formula:
BE = (BE/A) * A

Given that BE/A = 7.6 MeV and A = 240:
BE = (7.6 MeV) * 240
BE = 1824 MeV

Step 2: Calculation of binding energy of fragments:
The binding energy of each fragment can be calculated using the formula:
BE1 = (BE1/A1) * A1
BE2 = (BE2/A2) * A2

Given that BE1/A1 = BE2/A2 = 8.5 MeV and A1 = A2 = 120:
BE1 = (8.5 MeV) * 120
BE1 = 1020 MeV
BE2 = (8.5 MeV) * 120
BE2 = 1020 MeV

Step 3: Calculation of total gain in binding energy:
The total gain in binding energy is the difference between the binding energy of the unfragmented nucleus and the binding energy of the fragments:
Total gain in binding energy = BE - (BE1 + BE2)
Total gain in binding energy = 1824 MeV - (1020 MeV + 1020 MeV)
Total gain in binding energy = 1824 MeV - 2040 MeV
Total gain in binding energy = -216 MeV

Conclusion:
The total gain in the binding energy in the process is -216 MeV.
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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:a)0.9MeVb)9.4MeVc)804MeVd)216MeVCorrect answer is option 'D'. Can you explain this answer?
Question Description
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