An `alpha` – particle having kinetic energy 5 MeV falls on a Cu-foil. The shortest distance from the nucleus of Cu to which `alpha` - particle reaches is (Atomic no. of Cu = 29, K= `9xx10^9 Nm^2//C^2`)

Options

A. `2.35xx10^(-13)` m
B. `1.67 xx 10^(-14) m`
C. `5.98 xx 10^(-15)`
D. `1.67 xx 10^(-16)` m

Answer

B

Explanation

`K.E. = (K.Ze.2e)/(r)`
`r= (9xx10^9 xx 29xx 2xx1.6 xx 10^(-19)^2)/(5xx 1.6 xx 10^(-19) xx 10^(6))`
`r= (9xx10^9 xx 29 xx 2 xx 1.6 xx 10^(-19))/(5xx 10^6)`
`= 1.67 xx 10^(-14) m`