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Atoms and Nuclei question

2023 · 1 Feb · Shift 2 · Q68
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Atoms and Nuclei question

2023 · 1 Feb · Shift 2 · Q68

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
Nucleus A having Z=17Z=17Z=17 and equal number of protons and neutrons has 1.2 MeV1.2 ~\mathrm{MeV}1.2 MeV binding energy per nucleon. Another nucleus B\mathrm{B}B of Z=12Z=12Z=12 has total 26 nucleons and 1.8 MeV1.8 ~\mathrm{MeV}1.8 MeV binding energy per nucleons. The difference of binding energy of B\mathrm{B}B and A\mathrm{A}A will be ‾\underline{\hspace{2cm}}​MeV\mathrm{MeV}MeV.
Numerical answer
View written solutionFree

Correct answer: 6

  1. Find mass number of nucleus A

Given for nucleus AAA:

  • Atomic number Z=17Z = 17Z=17
  • Number of protons === number of neutrons

So, N=Z=17N = Z = 17N=Z=17 Hence mass number, AA=Z+N=17+17=34A_A = Z + N = 17 + 17 = 34AA​=Z+N=17+17=34

  1. Binding energy of nucleus A

Binding energy per nucleon of AAA is 1.2 MeV1.2\,\text{MeV}1.2MeV.

Therefore total binding energy of AAA is BEA=34×1.2=40.8 MeVBE_A = 34 \times 1.2 = 40.8\,\text{MeV}BEA​=34×1.2=40.8MeV

  1. Binding energy of nucleus B

Given for nucleus BBB:

  • Total nucleons =26= 26=26
  • Binding energy per nucleon =1.8 MeV= 1.8\,\text{MeV}=1.8MeV

So total binding energy of BBB is BEB=26×1.8=46.8 MeVBE_B = 26 \times 1.8 = 46.8\,\text{MeV}BEB​=26×1.8=46.8MeV

  1. Difference in binding energies

BEB−BEA=46.8−40.8=6.0 MeVBE_B - BE_A = 46.8 - 40.8 = 6.0\,\text{MeV}BEB​−BEA​=46.8−40.8=6.0MeV

So, the required difference is 6 MeV\boxed{6\,\text{MeV}}6MeV​

  1. Comparison with stored answer

Stored correct answer = 666

Our derived answer matches it.

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