With what velocity should an alpha particle travel towards the nucleus of a copper atom so as to arrive at a distance 10^{-13} m from the nucleus of a copper atom?

_{$\xb0$}which is equal to 10

^{-13}m here.

To arrive at a distance r

_{$\xb0$}from the nucleus, the kinetic energy of alpha particle should be equal to the potential energy of interaction of alpha particle with Cu nucleus.

or KE = PE

1/2 m

_{$\alpha $}v

_{$\alpha $}

^{2}= K q$\alpha $q

_{Cu}/r

_{$\xb0$}..................(1)

m

_{$\alpha $}= mass of $\alpha $-particle = 4 x 1.67 x 10

^{-27}kg.

K = 9 x 10

^{9}N$\xb7$m

^{2}/ C

^{2}

q

_{$\alpha $}= charge on alpha particle is +2 = 2 x (+1.6 x 10

^{-19}C)

q

_{Cu}= 29 x (+1.6 x 10

^{-19}C)

r

^{0}= 10

^{–13}m

Putting all the values in given equation (1) we get v$\alpha $

^{2}

${v}_{\alpha}^{2}=\frac{2\times 9\times {10}^{9}(2\times 1.6\times {10}^{-19})(29\times 1.6\times {10}^{-19})}{4\times 1.67\times {10}^{-27}\times {10}^{-13}}\phantom{\rule{0ex}{0ex}}=400.09\times {10}^{11}\phantom{\rule{0ex}{0ex}}=4000.9\times {10}^{10}\phantom{\rule{0ex}{0ex}}{v}_{\alpha}=\sqrt{4000.9\times {10}^{10}}\phantom{\rule{0ex}{0ex}}=63.25\times {10}^{5}m{s}^{-1}$

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