With what velocity should an `alpha`- particle travel towards the nucleus of a copper atom at a distance of `10^(-13)m` from the nucleus of the copper atom?

Explanation

As the alpha particle approaches the cu nucless , it decelectrates due to reqlision it and finaly its velocity becomes zero at point A (which is the turning point) After that , the `alpha` particle moves in the left derection (accelarating )
To arrive at a distance `v_(0)` from the nucless , the kinetic energy of alpha particle should be equal to the electrosstatic potential energy of it `i.e. KE = EPE`
`(1)/(2) m_(a)v_(a)^(2) = (Ka_(a)q_(N))/(v_(0))`
`m_(a)` = mass of `alpha ` particle `= 4 xx (1.67 xx 10^(-27) kg)`
Gv= velocity of `alpha` particle = ?
`K = 9 xx 10^(9) N m C^(-2)`
`q_(0)` = Charge on `alpha` particle `= 2 xx (+1.6 xx 10^(-19) C)`
`v_(0)` = Minilum distance of approach
`q_(N)` = cahrge on Cu nucless `= Ze = 29(+1.6 xx 10^(-19)C)`
d= Distance from necless `= 10^(-13) m`
`rArr Gv = sqrt((2Kq_(a)q_(N))/(m_(a)v_(0)))`
Substituting the given velues we get
`v_(a) = 6.325 xx 10^(6) ms^(-1)`