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Distance of closest approach of an alpha particle to a nucleus

This type of question, asked 3 times: 8 Apr 2024, 21 Jan 2026 S1, 22 Jan 2026 S1. All 3 questions below, oldest first.

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  1. Question 1

    8 Apr 2024

    30
    In an alpha particle scattering experiment distance of closest approach for the α\alpha particle is 4.5×10−14 m4.5 \times10^{- 14}\text{ }m. If target nucleus has atomic number 80 , then maximum velocity of α\alpha-particle is ___ ×105\times10^{5} m/sm/s approximately. (14πϵ0=9×109SI\left( \frac{1}{4\pi\epsilon_{0}}= 9 \times10^{9}SI \right. unit, mass of α\alpha particle == 6.72×10−27 kg6.72 \times10^{- 27}\text{ }kg )

  2. Question 2

    21 Jan 2026 S1

    13
    If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____\_\_\_\_ m . (Atomic number of gold =79= 79 and 14πϵ0=9×109\frac{1}{4\pi\epsilon_{0}}= 9 \times10^{9} in SI units)

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  3. Question 3

    22 Jan 2026 S1

    2
    7.9MeVα7.9MeV\alpha-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) ____\_\_\_\_ m. [14πϵo=9×109Nm2/C2\left\lbrack \frac{1}{4\pi\epsilon_{o}}= 9 \times10^{9}Nm^{2}/C^{2} \right. and electron charge  =1.6×10−19C]\left. \ = 1.6 \times10^{- 19}C \right\rbrack

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