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NVAEasyJEE 2023Atomic Mass & Binding Energy

JEE Physics 2023 Question with Solution

A nucleus with mass number 242242 and binding energy per nucleon as 7.6 MeV7.6 \, \text{MeV} breaks into two fragments each with mass number 121121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV8.1 \, \text{MeV}, the total gain in binding energy is:

Answer

Correct answer:121

Step-by-step solution

Standard Method

Given: Initial nucleus has mass number 242242 and binding energy per nucleon 7.6 MeV7.6 \, \text{MeV}. It breaks into two fragments, each of mass number 121121, with binding energy per nucleon 8.1 MeV8.1 \, \text{MeV}.

Find: The total gain in binding energy.

First calculate the initial total binding energy:

Binitial=242×7.6=1839.2 MeVB_{\text{initial}} = 242 \times 7.6 = 1839.2 \, \text{MeV}

Now calculate the final total binding energy of the two fragments:

Bfinal=2×(121×8.1)=2×980.1=1960.2 MeVB_{\text{final}} = 2 \times (121 \times 8.1) = 2 \times 980.1 = 1960.2 \, \text{MeV}

The gain in binding energy is the increase in total binding energy:

Gain=Bfinal−Binitial=1960.2−1839.2=121 MeV\text{Gain} = B_{\text{final}} - B_{\text{initial}} = 1960.2 - 1839.2 = 121 \, \text{MeV}

Therefore, the total gain in binding energy is 121 MeV121 \, \text{MeV}.

Binding Energy Interpretation

Given: A heavier nucleus splits into two equal fragments. The binding energy per nucleon increases from 7.6 MeV7.6 \, \text{MeV} to 8.1 MeV8.1 \, \text{MeV}.

Find: The total increase in binding energy after the reaction.

Binding energy gain indicates that the final nuclei are more stable than the initial nucleus. So we compare total binding energies before and after fission.

For the initial nucleus:

242 nucleons×7.6 MeV per nucleon=1839.2 MeV242 \text{ nucleons} \times 7.6 \, \text{MeV per nucleon} = 1839.2 \, \text{MeV}

For one fragment:

121×8.1=980.1 MeV121 \times 8.1 = 980.1 \, \text{MeV}

For two fragments:

2×980.1=1960.2 MeV2 \times 980.1 = 1960.2 \, \text{MeV}

Hence the increase is:

1960.2−1839.2=121 MeV1960.2 - 1839.2 = 121 \, \text{MeV}

Thus, the required numerical value is 121.

Common mistakes

  • Using only the difference in binding energy per nucleon, 8.1−7.68.1 - 7.6, as the final answer is wrong because the question asks for total gain in binding energy. Multiply by the total number of nucleons involved to get the full energy change.

  • Calculating the binding energy of only one fragment and stopping at 121×8.1121 \times 8.1 is incorrect because two identical fragments are produced. Include both fragments in the final total binding energy.

  • Subtracting in the wrong order, Binitial−BfinalB_{\text{initial}} - B_{\text{final}}, gives a negative value. Since the question asks for gain, compute Bfinal−BinitialB_{\text{final}} - B_{\text{initial}}.

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