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. Visible region
B. X - ray region
C. Ultra-violet region
D. Infra-red region

Answer

C

Explanation

From Einstein's photoelectric equation
`E=hv_(0)+1/2 mv^(2)=hv_(0)+eV_(0)`
where K.E. of the ejected electron `=eV_(0)=5eV` where `V_(0)` is the stopping potential and `hv_(0)=` threshold energy
`:.E=6.2 eV+5eV=11.2 eV`
`=11.2xx1.6xx10^(-19)J`
but `E=hv=(hc)/(lambda)`
`:.lambda=(hc)/(E)=(6.63xx10^(-34)xx3xx10^(8))/(11.2xx1.6xx10^(-19))`
`=1.11xx10^(-7)m`
`=1110Å`
This wavelength lies in the ultraviolet region.