Question

With what velocity should an $\alpha$-particle travel towards the nucleus of a copper atom so as to arrive at a distance $10^{-13}$ metre from the nucleus of the Cu-atom ? (Given $\in_{o}=8.85 \times 10^{-12} J ^{-1} C^{2} m ^{-1}$, atomic no. of $Cu =29$).

Answer 1

SOLUTION : When an $\alpha$-particle approaches a nucleus, it reverses its direction, i.e., scattering by $180^{\circ}$. At the instant of closest approach the initial kinetic energy of the $\alpha$-particle is entirely converted to electrostatic potential energy.

Potential cnergy (P.E.) $=\frac{\text { charge on the } \alpha-\text { particle } \times \text { charge on the } C u \text { atom }}{4 \pi \epsilon_{o} r}$

where $r=$ distance of closest approach $=10^{-13} m$

$\therefore \quad PE =\frac{2 e \times 29 e}{4 \pi \epsilon_{0} \times r}=\frac{58\left(1.60 \times 10^{-19}\right)^{2}}{4 \times 3 \cdot 1416 \times 8 \cdot 85 \times 10^{-12} \times 10^{-13}} J =1.33 \times 10^{-13} J$

Hence, K.E. of the $\alpha$-particle $= 1 . 3 3 \times 10^{-13} J$

Now K.E $=\frac{1}{2} \times m \times v^{2}$ for the $\alpha$-particle, mass of the $\alpha$-particle

$\therefore \quad v^{2}=\frac{2 KE }{m}=\frac{2 \times 1.33 \times 10^{-13}}{4 \times 1.67 \times 10^{-27}}=39 \cdot 82 \times 10^{12}$

$v=\left(39 \cdot 82 \times 10^{12}\right)^{\frac{1}{2}} ms ^{-1}=6 \cdot 31 \times 10^{6} ms ^{-1} \text {. }$