The graph between `1/lambda` and stopping potential `V_(0)` of two metals having work funcation `phi_(1)` and `phi_(2)` in an experiment of photoelectric effect is obtained as shown in the figure. Find out -

(i) Threshold wavelength of both metals
(ii) `phi_(1) : phi_(2)`
(iii) Which metal can emit photoelectrons with visible light ?

Answer

(i) `5000Å, 2000Å`,
(ii) `2:5` (ii) Metal-1

Explanation

(i) For metal 1 threshold wavelength is `lambda_(1)`
`1/lambda_(1)=0.002 nm^(-1)` or `lambda_(1)=500 nm=5000 Å`
For metal 2 threshold wavelength is `lambda_(2)`
`1/lambda_(2)=0.005 nm^(-1)` or `lambda_(2)=200 nm=2000 Å`
(ii) `phi=(hc)/lambda_(0)` or `phi prop 1/lambda_(0)" ":. phi_(1) : phi_(2)=2 : 5`
(iii) Metal 1 because `lambda_(1)` lies in visible wavelength range.