CBSE Class 12 Physics 2023: question 30(a)

Hints

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Step-by-step solution

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1. (i) **Distance of closest approach ($r_0$):** the minimum distance from the centre of the nucleus that an alpha particle reaches in a head-on approach. At this point its entire kinetic energy has been converted into electrostatic potential energy, so it momentarily stops and reverses. It is given by $r_0=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{K}$.

   1 mark

2. **Impact parameter ($b$):** the perpendicular distance of the initial velocity vector of the alpha particle from the centre of the nucleus. A small $b$ gives a large scattering angle; $b=0$ corresponds to a head-on collision.

   1 mark

3. (ii) Using $r_0=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{K}$ with $K=3.95\text{ MeV}=3.95\times 10^6\times 1.6\times 10^{-19}$ J:

   $r_0=\frac{9\times 10^9\times 2\times 79\times (1.6\times 10^{-19}{)}^2}{3.95\times 10^6\times 1.6\times 10^{-19}}$

   ½ mark

4. $r_0=5.76\times 10^{-14}\text{ m}=57.6\times 10^{-15}$ m, i.e. about 57.6 fm.

   ½ mark

Answer: (ii) r₀ ≈ 5.76 × 10⁻¹⁴ m (57.6 fm)

Revise the chapter

• Chapter: Atoms (notes, formulas, where marks are lost and every past question)
• Topic: Alpha-Particle Scattering and Rutherford's Model