Question

The velocity with which an $\alpha$ - particle travel towards the nucleus of a copper atom so as to arrive at a distance ${10}^{-13}$ meter from the nucleus of the copper atom is:

• A. $6.3\times {10}^{6}mse{c}^{-1}$
• B. $6.8\times {10}^{6}mse{c}^{-1}$
• C. $7.1\times {10}^{6}mse{c}^{-1}$
• D. $7.4\times {10}^{6}mse{c}^{-1}$

Answer 1

The correct option is A $6.3\times {10}^{6}mse{c}^{-1}$

$K.E=P.E$

$\frac{1}{2}m{u}^2=\frac{kZ{Z}^{\prime }}{r}$

$\frac{1}{2}\times 6.696\times {10}^{-27}{u}^2=\frac{9\times {10}^9\times 29\times 2(1.602\times {10}^{-19}{)}^2}{{10}^{-13}}$

$u=6.32\times {10}^{6}mse{c}^{-1}$