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

We know that
`hf_1=phi+(1)/(2)mv_1^(2)` and `(1)/(2)mv_1^2=eV_1`
`phi=hf_1-eV_1=(hc)/(lamda_1)-eV=((6.6xx10^(-34))xx(3xx10^(8)))/(4950xx10^(-10))-(1.6xx10^(-19))(0.6)`
`=3.04xx10^(-19)V`,`hf_2=phi+eV_2`
`(hc)/(lamda_2)=3.04xx10^(-19)+(1.6xx10^(-19))xx1.1=4.8xx10^(-19)`
`implieslamda_2=((6.6xx10^(34))xx(3xx10^(8)))/(4.8xx10^(-19))=4125A`