In a discharge tube, which contains argon at low pressure, V potential difference is applied between two of its electrodes. Ionisation energy of argon atom is 15.6 eV. Separation between electrodes is `4.0 xx 10^(-2)` m and average distance that electron travels between two successive collisions with argon atoms is `8 xx 10^(-5)` m. Estimate the minimum value of V such that collision of electron may cause ionisation of argon atoms.

Explanation

Electric field between the plates can be written as follows :
E=V/d , where `d(=4.0xx10^(-2)m)` is separation between the plates.
If `lambda(=8xx10^(-5)m)` is distance between two successive collisions then energy acquired by the electron can be written as follows :
Energy acquired by electron before collision `=eElambda=eVlambda//d`
This energy must be equal to 15.6 eV to cause ionisation of argon atom.
`(eVlambda)/d=15.6xxe`
`V=(15.6xxd)/lambda=(15.6xx4xx10^(-2))/(8xx10^5)`
`=7.8xx10^3` V = 7.8 kV