An α -particle after passing through potential difference of V volt...

Given $KE$ of $\alpha$-particle $= 2eV$

$r = \dfrac{\left(Ze\right)\left(e\right)}{4\pi{ε}_{0} \cdot \left(KE\right)}$

$= \dfrac{2Ze \times 9 \times {10}^{9}}{2V}$

$⇒ r = \dfrac{2 \times Z \times 16 \times {10}^{- 19} \times 9 \times {10}^{9}}{2V}$

$= 14. 4 \cdot \dfrac{Z}{V} Å$