Question

'$\alpha$ particle' 3.6 MeV are fired towards nucleus ${}_Z^{A}X$, at point of closest separation distance between '$\alpha$ particle' and 'X' is $1.6\times {10}^{-14}$ m. Calculate the atomic number of 'X'
[Given : $1/4\pi {ϵ}_0=9\times {10}^9$ in S. I. units]

Answer 1

When $\alpha -$ particle is fired the interaction between both nucleus decide the closest separation distance.

Here, $\frac{1}{4\Pi {\in }_o}\times \left(\frac{q_1{q}_2}{r}\right)=3.6\times {10}^6\times (1.6\times {10}^{-19})J$

$4$ potential energy

$\Rightarrow 9\times {10}^9\times \frac{(Z\times 1.6\times {10}^{-19})(2\times 1.6\times {10}^{-19})}{1.6\times {10}^{-14}}=(3.6\times {10}^{+6})\times (1.6\times {10}^{-19})$

$\Rightarrow Z\times 2\times 9\times {10}^4=3.6\times {10}^6$

$\Rightarrow Z\times 18=360$

$\Rightarrow Z=20$

Atomic number of $X=20$