Question

Find the distance of closest approach when an $\alpha$ particle is projected towards the nucleus of silver atom having speed $v$.

$(massofparticle=m_{\alpha },atomicnumberofAg=47)$

Answer 1

$E_{\alpha }=q{V}_E=q\left(\frac{1}{4\Pi {\epsilon }_o}\frac{Q}{d_o}\right)$

$\Rightarrow d_o=\frac{1}{4\Pi {\epsilon }_o}\frac{qQ}{E\alpha }$

$\Rightarrow E_{\alpha }=K.E=\frac{1}{2}m\alpha V^2$

$q=+2e,Q=Ze=47e$

$\Rightarrow d_o=\frac{1}{4\Pi {\epsilon }_o}\times \frac{2\times 47\times e^2}{\frac{1}{2}{m}_{\alpha }{V}^2}$

$=\frac{4.331\times {10}^{-26}}{m_{\alpha }{V}^2}$