Question

With what velocity should $\alpha$- Particle travel towards the neucleus of a Cu atom so as to arrive at a distance of ${10}^{-13}m$.

Answer 1

For reaching $Cu$ atom, $kE=pE$

$\Rightarrow \frac{1}{2}{m}_{\alpha }{V}_{\alpha }^2=\frac{k{q}_{\alpha }{q}_{Cu}}{r}$

Now, $m_{\alpha }=4\times 1.67\times {10}^{-27}kg=6.68\times {10}^{-27}kg$

$k=9\times {10}^{9}N{m}^2/c^2$

$q_{\alpha }=2(1.6\times {10}^{-19})C=3.2\times {10}^{-19}$

$r={10}^{-13}m$

$q_{Cu}=29(1.6\times {10}^{-19})C=4.64\times {10}^{-18}C$

$\therefore \frac{1}{2}(6.68\times {10}^{-27})(V_{\alpha }{)}^2=\frac{(9\times {10}^9)(3.2\times {10}^{-19})(4.64\times {10}^{-18})}{{10}^{-13}}$

$\Rightarrow V_{\alpha }=63.25m/s$