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Atoms and Nuclei question

2023 · 1 Feb · Shift 1 · Q59
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Atoms and Nuclei question

2023 · 1 Feb · Shift 1 · Q59

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
The mass of proton, neutron and helium nucleus are respectively 1.0073 u,1.0087 u1.0073~u,1.0087~u1.0073 u,1.0087 u and 4.0015 u4.0015~u4.0015 u. The binding energy of helium nucleus is :
  1. A
    28.4 MeV28.4~\mathrm{MeV}28.4 MeV
  2. B
    56.8 MeV56.8~\mathrm{MeV}56.8 MeV
  3. C
    7.1 MeV7.1~\mathrm{MeV}7.1 MeV
  4. D
    14.2 MeV14.2~\mathrm{MeV}14.2 MeV
View written solutionFree

Correct answer: A

  1. Write the nucleus composition

A helium nucleus (24He{}^4_2\mathrm{He}24​He) contains:

  • 222 protons
  • 222 neutrons

So, the total mass of the separated nucleons is mseparate=2mp+2mnm_{\text{separate}} = 2m_p + 2m_nmseparate​=2mp​+2mn​

  1. Substitute the given masses

Given: mp=1.0073 u,mn=1.0087 um_p = 1.0073\,u, \quad m_n = 1.0087\,ump​=1.0073u,mn​=1.0087u mHe nucleus=4.0015 um_{\text{He nucleus}} = 4.0015\,umHe nucleus​=4.0015u

Hence, mseparate=2(1.0073)+2(1.0087)m_{\text{separate}} = 2(1.0073) + 2(1.0087)mseparate​=2(1.0073)+2(1.0087) =2.0146+2.0174= 2.0146 + 2.0174=2.0146+2.0174 =4.0320 u= 4.0320\,u=4.0320u

  1. Find the mass defect

Mass defect is Δm=mseparate−mnucleus\Delta m = m_{\text{separate}} - m_{\text{nucleus}}Δm=mseparate​−mnucleus​ =4.0320−4.0015= 4.0320 - 4.0015=4.0320−4.0015 =0.0305 u= 0.0305\,u=0.0305u

  1. Convert mass defect into binding energy

Using 1 u=931.5 MeV1\,u = 931.5\,\text{MeV}1u=931.5MeV

Therefore, B.E.=Δm×931.5B.E. = \Delta m \times 931.5B.E.=Δm×931.5 =0.0305×931.5= 0.0305 \times 931.5=0.0305×931.5 ≈28.41 MeV\approx 28.41\,\text{MeV}≈28.41MeV

  1. Match with the options

28.4 MeV\boxed{28.4\,\text{MeV}}28.4MeV​

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

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