An alpha particle after passing through a potential difference of `2xx10^(6)` volt falls ona silver foil. The atomic number of silver is 47. Calculate (i) the K.E. of the alpha-particle at the time of falling on the foil. (ii) K.E. of the `alpha`-particle at a distance of `5xx10^(-14)m` from the nucleus, (iii) the shortest distance from the nucleus of silver to which the `alpha`-particle reaches.

Answer

`6.4xx10^(-13)J,2.1xx10^(-13),3.4xx10^(-14)m`

Explanation

(i) K.E. of a-particlue for Al for
K.E.=`q.V=6.4xx10^(-13)J" " (q=2xx1.6xx10^(-19))`
(ii) `P.E=(kq_(1)q_(2))/(r)=(9xx10^(9)xx2xx1.6xx10^(19)xx47xx1.6xx10^(-19))/(5xx10^(-14))`
`=4.33xx10^(-13)J`
Net K.E.=K.l-P.E.`=(6.4-4.3)XX10^(-13)J`
`2.07xx10^(-13)J`
(iii) K.E=P.E.
`6.4xx10^(-13)=(2.16xx10^(-26))/(r)`
`r=3.4xx10^(-14)m`