Question

(a) (K.E.)_Initial = (P.E.)_{at distance of closest approach}

$\text{Or 4.0 MeV = K.}\frac{q_{1}q_{2}}{r} $$$\text{Or, 4} \times \text{1}\text{0}^{6} \times 1.6 \times 10^{- 19} = 9 \times 10^{9} \times \frac{\left( 2 \times 1.6 \times 10^{- 19} \right) \times \left( 50 \times 1.6 \times 10^{- 19} \right)}{r}$$

∴ Distance of closest approach, r = 3.6 × 10^–14 m

(b) $P.E. = K.\frac{q_{1}q_{2}}{r} = 9 \times 10^{9} \times \frac{\left( 2 \times 1.6 \times 10^{- 19} \right) \times \left( 50 \times 1.6 \times 10^{- 19} \right)}{9 \times 10^{- 14}}$

$= 10^{25} \times \left( 1.6 \times 10^{- 19} \right)^{2}\text{ J =}\frac{10^{25} \times \left( 1.6 \times 10^{- 19} \right)^{2}}{1.6 \times 10^{- 19}}\text{ eV = 1.6 MeV}$

(c) $P.E. = K.\frac{q_{1}q_{2}}{r} = \frac{9 \times 10^{9} \times \left( 2 \times 1.6 \times 10^{- 19} \right) \times \left( 50 \times 1.6 \times 10^{- 19} \right)}{4.5 \times 10^{- 14} \times \left( 1.6 \times 10^{- 19} \times 10^{6} \right)} = 3.2\text{ MeV}$

∴ K.E. of $\alpha -$particle at this distance = 4.0 – 3.2 = 0.8 MeV