The maximum KE of photoelectrons emitted from a certain metallic surface is 30 eV when monochromatic radiation of wavelength `lamda` falls on it. When the same surface is illuminated with light of wavelength `2lamda`, the minimum KE of photoelectrons is found to be 10 eV. (a) Calculate the wavelength `lamda` and (b) determine the maximum wavelength of incident radiation for which photoelectric emission is possible.

Explanation

According to Einstein's photoelectric equation,
`(hc)/(lamda)=W+K_(max)`
Here, `(hc)/(lamda)=W+30eV` and `(hc)/(2lamda)=W+10eV`
`(hc)/(lamda)-(hc)/(2lamda)=(hc)/(2lamda)=20eV`
`lamda=(hc)/(40eV)=((6.62xx10^(-34))(3xx10^(8)))/(40xx(1.6xx10^(19)))=310.3`
Now, `W=(hc)/(lamda)-K_(max)=(40-30)eV=10eV`
`lamda_0=((6.62xx10^(-34))(3xx10^(8)))/(10xx(1.6xxxx10^(19)))=1241 A`