An `alpha`-particle after passing through a potential difference of `2 xx 10^(6) V` falls on a silver foil. The atomic number of silver is 47. Calculaye (i) the kinetic energy of the `alpha`-particle at the time of falling on the foil (ii) the kinetic energy of the `alpha`-particle at a distance of `5 xx 10^(-14)` m from the nucleus and (iii) the shortest distance from the nucleus of silver to which the `alpha`-particle reaches.

Answer

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

Explanation

(a) `kE=qV=2xx1.6^(-19)xx2xx10^(6)=6.4xx10^(-13)J`
(b) At distance `d=5xx10^(-14)m` let K.E. is x J and
`PE=(kq_(1)q_(2))/(d)=(9xx10^(9)xx2xx1.6xx10^(-19)xx47xx1.6xx10^(-19))/(5xx10^(-14))`
`PE=4.33xx10^(-13)J`
By energy conservation :
`6.4xx10^(-13)=x+4.33xx10^(-13)`
`x=2.06xx10^(-13)J, " "kE=PE`
`6.4xx10^(-13)=(9xx10^(9)xx2xx47xx(1.6xx10^(-19))^(2))/(d)`
`implies d=3.384xx10^(-14)m`