Energy corresponding to threshold frequency of metal is 6.2 eV.For given radiation stopping potential is 5 V then incident radiation will be in …region.

Options

A. ultraviolet
B. visible region
C. infrared region
D. X-rays region

Answer

A

Explanation

Work function `phi_(0)`=6.2 eV
Stopping potential `V_(0)`=5V
Now `eV_(0)=hf-phi_(0)`
`eV_(0)+phi_(0)=(hc)/(lambda)`
`therefore lambda =(hc)/(eV_(0)+phi_(0))`
`=(6.62xx10^(-34)Jsxx3xx10^(8)ms^(-1))/(1.6xx10^(-19)xx5J+6.2xx1.6xx10^(-19)J)`
`=(6.62xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx(5+6.2))`
`=(19.86xx10^(-26))/(17.92xx10^(-19))`
`=1.10xx10^(-7)`m
From this it can be said that wavelength of incident radiation is in ultraviolet region.