Work function of material is 4.0 eV. Maximum wavelength by which photoelectric emission can be obtained is ……….

Options

A. 540 nm
B. 400 nm
C. 310 nm
D. 220 nm

Answer

C

Explanation

`phi=4eV=4xx1.6xx10^(-19)J`
`therefore phi=(hc)/(lambda_(0))`
Maximum wavelength `lambda`=?
`therefore lambda_(0)=(hc)/(phi)=(6.62xx10^(-34)xx3xx10^(8))/(4xx1.6xx10^(-19))`
`=3.103xx10^(-7)m`
`310.3xx10^(-9)m ~~310 nm`