(a) (i) Differentiate between 'distance of closest approach' and 'impact parameter'. (ii) Determine the distance of closest approach when an alpha…

Distance of closest approach: It is the minimum distance of α-particle from the centre of nucleus at which its total kinetic energy gets converted into electrostatic potential energy.

Alternatively: Distance of closest approach

\[r_o = \frac{1}{4\pi \varepsilon_o}\frac{2Ze^2}{E_K} \]

Impact parameter: Perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus.

\[r_o = \frac{1}{4\pi \varepsilon_o}\frac{2Ze^2}{E_K} \]

\[r_o = \frac{9\times 10^{9}\times 2\times 79\times (1.6\times 10^{-19})^{2}}{3.95\times 1.6\times 10^{-13}} \]

\[= 57.6\times 10^{-15}\, m \]

OR(b) (i) (1) An electron in an atom revolves in certain stable orbit without the emission of radiant energy.

The electron revolves around the nucleus only in those orbits for which the angular momentum is integral multiple of $\displaystyle \dfrac{h}{2\pi}$ where h is the Planck's constant.

When an electron makes a transition from one of its specified non-radiating orbits to another of lower energy orbit a photon is emitted having energy equal to the energy difference between the initial and final states.

Angular momentum $\displaystyle = \dfrac{nh}{2\pi}$

For n = $\displaystyle 2$

Angular momentum $\displaystyle = \dfrac{2h}{2\pi}$

\[= \frac{6.63\times 10^{-34}}{\pi} \]

\[= 2.1\times 10^{-34}\ kg\, m^{2} s^{-1} \]