MCQEasyJEE 2024Atomic Mass & Binding Energy

JEE Physics 2024 Question with Solution

The atomic mass of 612C^{12}_{6}\text{C} is 12.000000u12.000000 \, \text{u} and that of 613C^{13}_{6}\text{C} is 13.003354u13.003354 \, \text{u}. The required energy to remove a neutron from 613C^{13}_{6}\text{C}, if the mass of the neutron is 1.008665u1.008665 \, \text{u}, will be:

  • A

    62.5MeV62.5 \, \text{MeV}

  • B

    6.25MeV6.25 \, \text{MeV}

  • C

    4.95MeV4.95 \, \text{MeV}

  • D

    49.5MeV49.5 \, \text{MeV}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Atomic mass of 612C^{12}_{6}\text{C} is 12.000000u12.000000 \, \text{u}, atomic mass of 613C^{13}_{6}\text{C} is 13.003354u13.003354 \, \text{u}, and neutron mass is 1.008665u1.008665 \, \text{u}.

Find: The energy required to remove one neutron from 613C^{13}_{6}\text{C}.

The process is

613C612C+neutron^{13}_{6}\text{C} \rightarrow ^{12}_{6}\text{C} + \text{neutron}

The mass difference is

Δm=(12.000000+1.008665)13.003354\Delta m = \left(12.000000 + 1.008665\right) - 13.003354 Δm=13.00866513.003354=0.005311u\Delta m = 13.008665 - 13.003354 = 0.005311 \, \text{u}

Now convert this mass defect into energy using 1u=931.5MeV1 \, \text{u} = 931.5 \, \text{MeV}:

E=Δm×931.5E = \Delta m \times 931.5 E=0.005311×931.54.95MeVE = 0.005311 \times 931.5 \approx 4.95 \, \text{MeV}

Therefore, the energy required to remove a neutron is 4.95MeV4.95 \, \text{MeV}. The correct option is C.

The solution also states this as the binding energy difference between 613C^{13}_{6}\text{C} and 612C^{12}_{6}\text{C}, which gives the same result.

Binding Energy Difference Method

Given: The nuclei are 612C^{12}_{6}\text{C} and 613C^{13}_{6}\text{C}.

Find: Separation energy of one neutron from 613C^{13}_{6}\text{C}.

Using the idea from the solution, the neutron separation energy is the difference in binding energies:

ΔBE=BE613CBE612C\Delta \text{BE} = \text{BE}_{^{13}_{6}\text{C}} - \text{BE}_{^{12}_{6}\text{C}}

This is equivalent to comparing the masses of the parent nucleus and the products:

ΔM=M(613C)M(612C)mn\Delta M = M\left(^{13}_{6}\text{C}\right) - M\left(^{12}_{6}\text{C}\right) - m_n ΔM=13.00335412.0000001.008665=0.005311u\Delta M = 13.003354 - 12.000000 - 1.008665 = -0.005311 \, \text{u}

Taking the magnitude for required energy,

ΔM=0.005311u|\Delta M| = 0.005311 \, \text{u}

Hence,

ΔE=ΔM×931.5=0.005311×931.54.950MeV\Delta E = |\Delta M| \times 931.5 = 0.005311 \times 931.5 \approx 4.950 \, \text{MeV}

Therefore, the required neutron removal energy is 4.95MeV4.95 \, \text{MeV} and the correct option is C.

Common mistakes

  • Using 13.00335412.0000001.00866513.003354 - 12.000000 - 1.008665 and directly calling the negative sign the final energy is incorrect. The negative sign only indicates mass decreases in the bookkeeping; the required energy is the magnitude of the mass defect times 931.5MeV/u931.5 \, \text{MeV/u}.

  • Subtracting masses in the wrong reaction direction is a common mistake. Write the reaction first as 613C612C+neutron^{13}_{6}\text{C} \rightarrow ^{12}_{6}\text{C} + \text{neutron}, then compute product mass minus reactant mass for the energy required.

  • Confusing binding energy of the whole nucleus with neutron separation energy is wrong. Here only one neutron is being removed, so you need the mass difference between 613C^{13}_{6}\text{C} and 612C+n^{12}_{6}\text{C} + n, not the total binding energy of 613C^{13}_{6}\text{C}.

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