1.

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 :

Answer»

`(10.39)/(1.5 XX 10^(6))m`
`10.39 xx 1.5 xx 10^(6)m`
`10.39 xx 1.6 xx 10^(-19)m`
`(10.39 xx 1.6 xx 10^(-19))/(1.5 xx 10^(6))m`

Solution :`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`


Discussion

No Comment Found

Related InterviewSolutions