The ionisation potential of mercury is 10.39 volt. To gain energy sufficient enough to ionise mercury, an electron must travel in an electric field of `1.5 xx 10^6 Vm^(-1)` at distance of :

Options

A. `(10.39)/(1.5 xx 10^(6))m`
B. `10.39 xx 1.5 xx 10^(6)m`
C. `10.39 xx 1.6 xx 10^(-19)m`
D. `(10.39 xx 1.6 xx 10^(-19))/(1.5 xx 10^(6))m`

Answer

A

Explanation

`10.39eV =1.5 xx 10^(6) xx d xx 1.6 xx 10^(-19) (therefore w=eEd)`
`10.39 xx 1.6 xx 10^(-19) =1.5 xx 10^(6) xx d xx 1.6 xx 10^(-19)`
`d=(10.39)/(1.5 xx 10^(6))m`