`alpha`-particles of enegry `400 KeV` are boumbardel on nucleus of `._(82)Pb` . In scattering of `alpha`-particles, it minimum distance from nucleus will be

Options

A. `0.59 nm`
B. `0.59 Å`
C. `0.59 p m`
D. `0.59 p m`

Answer

D

Explanation

Suppose closet distance is `r`, according to conservation of energy.
`400 xx 10^(3) xx 1.6 xx 10^(-19) = 9 xx 10^(9) ((ze)(2e))/(r )`
`rArr 6.4 xx 10^(-14)`
`= (9 xx 10^(9) xx (82 xx 1.6 xx 10^(-19))xx (2 xx 1.6 xx 10^(-19)))/(r )`
`rArr r = 5.9 xx 10^(-13) m = 0.59p m`