With what velocity should an α–particle travel towards the nucleus of copper atoms so as to arrive a

Question

With what velocity should an α–particle travel towards the nucleus of copper atoms so as to arrive at a distance 10–12 m from the nucleus of the copper atom? (4.8 ×$\sqrt{29 	imes 60}$ = 200, NA = 6 × 1023, e = 1.6 × 10−19 C)

Accepted Answer

$\frac{1}{2}mv^{2} = K.\frac{q_{1}q_{2}}{r}$

$\text{Or, }\frac{1}{2} \times \left( \frac{4 \times 10^{- 3}}{6 \times 10^{23}} \right) \times V^{2} = 9 \times 10^{9} \times \frac{\left( 2 \times 1.6 \times 10^{- 19} \right) \times \left( 29 \times 1.6 \times 10^{- 19} \right)}{10^{- 12}} $$$\therefore v = 2 \times 10^{6}\text{ m/s}$$

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