When light of wavelength 400nm is incident on the cathode of photocell, the stopping potential recorded is 6V. If the wavelength of the incident light is to 600nm, calculate the new stopping potential. [Given `h=6.6xx10^(-34) Js, c=3xx10^(8)m//s , e=1.6xx10^(-19)C`]

Options

A. `4.97 V`
B. `4.76V`
C. `4.56V`
D. `4.14V`

Answer

d

Explanation

As `eV=(hc)/lambda-phi_(0)`
case (i) `exx6=(1240 eVnm)/(400nm)-phi_(0).........(i)`
`=(3.1-phi_(0))eV`
Case (ii) `exxV'=(1240 eV nm)/((400+600)nm)-phi_(0)......(ii)`
`=(1.240-phi_(0))eV`
From (i) and (ii)
`6-V'=3.10-1.25=1.86`
`V'=6-1.86=4.14 V`