Q: α-particles of energy 400 KeV are bombarded on nucleus of _{82}Pb. In scattering of an α-particles, its minimum distance from nucleus will be

(a) 0.59 nm

(b) 0.59 Å

(c) 5.9 pm

(d) 0.59 pm

**Click to See Answer : **

Ans: (d)

Sol: Let r be the minimum distance from the nucleus i.e. distance of closest approach

At the distance of closest approach K.E is converted into P.E

$\displaystyle \frac{1}{4\pi \epsilon_0}.\frac{(Z e)(2 e)}{r} = 400 KeV $

$\displaystyle 9 \times 10^9 .\frac{(82 \times 1.6\times 10^{-19}) (2 \times 1.6\times 10^{-19})}{r} = 400\times 10^3 \times 1.6 \times 10^{-19} $

On solving we get ,

r = 0.59 pm