Question

An alpha particle after passing through a potential difference of $2\times {10}^{6}volt$ falls on a silver foil. The atomic number of silver is $47$. Calculate $\left(I\right)$ the $K.E$ of the alpha-particle at the time of falling on the foil. $\left(II\right)K.E$ of the alpha-particle at a distance of $5\times {10}^{-14}m$ from the nucleus, $\left(III\right)$ the shortest distance from the nucleus of silver to which the alpha-particle reaches.

Answer 1

Potential difference $\left(v\right)=2\times {10}^{6}volts$

Charge of $\propto$ particle $=2\times e^{-}$

Charge of silver $=47\times e^{-}$

$\left(i\right)K.E$ of $\alpha -$particle $=2e^{-}v$

$=2\times 1.602\times {10}^{-19}\times 2\times {10}^6$

$=6.4\times {10}^{-13}J$

$\left(ii\right){\left(KE\right)}_1={\left(KE\right)}_2+{\left(PE\right)}_2$

$\therefore {\left(KE\right)}_2={\left(KE\right)}_1-{\left(PE\right)}_2$

$=(6.4\times {10}^{-13})-\left(\frac{k{q}_1{q}_2}{r}\right)$ $(q_1=Chargeon\alpha -particle{q}_2=ChargeonAg=47\times e^{-})$

$\therefore {\left(KE\right)}_2=2.1\times {10}^{-13}J$