An `alpha`-particle is accelerated through a potential of 1MV and falls on a silver foil (Z=47).
Calculate : (i)Kinetic energy of the `alpha`-particle when it falls on the foil .
(ii)Kinetic energy of the `alpha`-particle when it is `6.9xx10^(-14)` m distant from the nucleus .
(iii)Distance of closest approach to the nucleus .

Explanation

To solve the problem step by step, we will address each part of the question systematically.

### Given Data:
- Potential difference (V) = 1 MV = \(1 \times 10^6\) V
- Charge of alpha particle (Q) = \(2e\) where \(e = 1.6 \times 10^{-19}\) C
- Atomic number of silver (Z) = 47
- Distance from the nucleus (r) = \(6.9 \times 10^{-14}\) m
- Coulomb's constant (k) = \(9 \times 10^9 \, \text{N m}^2/\text{C}^2\)
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