The threshold frequency for a metallic surface corresponds to an energy of `6.2eV` and the stopping potential for a radiation incident on this surface is `5 V` . The incident radiation lies in

Options

A. ultra-violet region
B. infra-red region
C. visible region
D. X-ray region

Answer

A

Explanation

`phi = 6.2 eV = 6.2 xx 1.6 xx 10^(-19) J`
V = 5 volt
`(hc)/(lamda) - phi = eV_0 rArr lamda = (hc)/(phi + eV_0)`
` = (6.6 xx 10^(-34) xx 3 xx 10^8)/(1.6 xx 10^(-19) (6.2 + 5)) ~~ 10^(-7) m`
This range lies in ultra violet range.