The work function of a surface of a photosensitive material is `6.2 eV`. The wavelength of the incident radiation for which the stopping potential is `5 V` lies in the

Options

A. Infrared region
B. X-ray region
C. Ultraviolet regi
D. Visible region

Answer

C

Explanation

`E_("incident")=W+K_("max"),K_("max")=eV_(0)`
`hv=hv_(0)+eV_(0)," stopping potential "=V_(0)`
`(hc)/(lambda)=6.2e+5e`
`lambda=(hxxc)/(11.2xx1.6xx10^(-19))=(6.63xx10^(-34)xx3xx10^(8))/(11.2xx1.6xx10^(-19))`
`=1.1xx10^(-7)m`