When a surface is irradiated with light of wavelength 4950 , a photocurrent appears which vanishes if a retarding potential greater than 0.6 V is applied across the phototube. When different source of light is used, it is found that the critical retarding potential is changed to 1.1 V. Find the work function of the emitted surface and the wavelength of the second source.
Given, wavelength of light used, λ=4950˙A
Stopping potential, $V_{o} =0.6V$
According to Einstein potential equation we have;
12mv2=eVo=hv−Vo........ (1)
ie, eVo=hcλ−ϕo
Where, ϕo = work function
Vo = stopping potential
For the first source
λ1=4950˙A=4950×10−10
Vo=0.6v
Putting these values in equation (1), we get
1.6×10−19×0.6=6.6×10−34×3×108495×10−9−ϕo
ϕo=3.04×10−19J
=3.04×10−191.6×10−19eV
=1.9eV
Let λ2 be the wavelength of the second source.
Stopping potential Vo =1.1 v (given)
Therefore,
1.6×10−19×1.1=6.6×10−34×3×108λ2−3.04×10−19J
1.76×10−19=19.8×10−27λ2−3.04×10−19
19.8×10−26λ2=4.8×10−19
∴λ2=19.8×10−264.8×10−19=4.125×10−7m
=4125˙A