When a surface is irradiated with light of wavelength `4950A`, a photocurrent appears which vanishes if a retarding potential greater than `0.6V` is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to `1.1V`. Find the work function of the emitting surface and the wavelength of the second source.

Explanation

a. `hupsilon_1=W+(1)/(2)mv_1^2`
and `(1)/(2)mv_1^2=eV_1`
`W=hupsilon_1-eV_1=(hc)/(lamda_1)-eV_1`
`=((6.6xx10^(-34))xx(3xx10^(8)))/(4950xx10^(-10))-(1.6xx10^(-19))(0.6)`
`=1.9eV=1.9xx(1.6xx10^(-19))V`
`=3.04xx10^(-19)V`
b. `hupsilon_(2)=W+eV_(2)`
or `(hc)/(lamda_(2))=3.04xx10^(-19)+(1.6xx10^(-19))xx1.1`
`=4.8xx10^(-19)`
`lamda_(2)=(hc)/(4.8xx10^(-19))=((6.6xx10^(-34))xx(3xx10^(8)))/(4.8xx10^(-19))=4125 A`
c. Since the magnetic field does not change the speed of ejected electrons, there will be no change in the stopping potential.