Binding | Modern Physics-II(Nuclear Physics) | JEE MAIN_2026

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JEE MAIN_2026

02-04-26_S2

Question

The binding energy per nucleon of ${ }_{83}^{209} \mathrm{Bi}$ is $\_\_\_\_$ MeV . [Take $\mathrm{m}\left({ }_{83}^{209} \mathrm{Bi}\right)=208.980388 \mathrm{u}, \mathrm{m}_{\mathrm{p}}=1.007825 \mathrm{u}, \mathrm{m}_{\mathrm{n}}=1.008665 \mathrm{u}, 1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^2$ ]

Select the correct option:

A

7.48

B

7.84

C

8.79

D

6.94

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

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