An `alpha` - particle after passing through a potential difference of V volts collides with a nucleus . If the atomic number of the nucleus is Z then the distance of closest approach of `alpha` – particle to the nucleus will be

Options

A. `14.4.(Z)/(V)`
B. `14.4(Z)/(V)m`
C. `14.4(V)/(Z)m`
D. `14.4(V)/(Z)`

Answer

a

Explanation

Given KE of `alpha`- particle =2eV
`r=((Ze)(e ))/(4piepsilon_(0)*(KE))`
`=(2Zexx9xx10^(9))/(2V)`
`rArrr=(2xxZxx1.6xx10^(-19)xx9xx10^(9))/(2V)`
`=14.4*(Z)/(V)Å`