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MCQEasyJEE 2026Rutherford's Model of Atom

JEE Physics 2026 Question with Solution

A 7.9 MeV7.9\, \text{MeV} α\alpha-particle scatters from a target material of atomic number 7979. From the given data, the estimated diameter of the nuclei of the target material is (approximately) _____ m\text{m}. [14πε0=9×109 N m2/C2 and electron charge =1.6×10−19 C]\left[\frac{1}{4\pi\varepsilon_0}=9\times10^9\, \text{N m}^2/\text{C}^2 \text{ and electron charge } =1.6\times10^{-19}\, \text{C}\right]

  • A

    1.69×10−121.69\times10^{-12}

  • B

    1.44×10−131.44\times10^{-13}

  • C

    2.88×10−142.88\times10^{-14}

  • D

    5.76×10−145.76\times10^{-14}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The incident particle is an α\alpha-particle with energy E=7.9 MeVE=7.9\, \text{MeV} and the target atomic number is Z=79Z=79.

Find: The estimated diameter of the target nucleus.

Use the distance of closest approach formula for Rutherford scattering:

d=14πε02Ze2Ed=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{E}

where e=1.6×10−19 Ce=1.6\times10^{-19}\, \text{C} and

E=7.9×106×1.6×10−19 JE=7.9\times10^6\times1.6\times10^{-19}\, \text{J}

Substituting the values:

d=9×109×2×79×(1.6×10−19)27.9×106×1.6×10−19d=9\times10^9\times \frac{2\times79\times(1.6\times10^{-19})^2}{7.9\times10^6\times1.6\times10^{-19}}

Therefore,

d≈2.88×10−14 md\approx 2.88\times10^{-14}\, \text{m}

Therefore, the estimated diameter of the nucleus is 2.88×10−14 m2.88\times10^{-14}\, \text{m} and the correct option is C.

Common mistakes

  • Using the atomic number 7979 directly as the projectile charge is incorrect. The α\alpha-particle has charge +2e+2e, which is why the factor 2Ze22Ze^2 appears in the formula. Use the projectile charge and target charge together.

  • Leaving the kinetic energy in MeV\text{MeV} while using SI values of 14πε0\frac{1}{4\pi\varepsilon_0} and ee is incorrect. Convert 7.9 MeV7.9\, \text{MeV} to joules before substitution so that the distance comes out in metres.

  • Confusing radius with diameter leads to formula errors. In this solution, the given expression directly yields the required closest-approach distance used as the estimated nuclear diameter, so do not divide or multiply by 22 again without justification.

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