Surface of certain metal is first illuminated with light of wavelength `lambda_(1)=350` nm and then, by light of wavelength `lambda_(2)=540` nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to :
(Energy of photon `=(1240)/(lambda("in nm"))Ev`

Explanation

From Einstein.s photoelectric equation
`(hc)/(1_(1))=f+1/2m(2v)^(2)`……I
and `(hc)/(1_(2))=f+1/2mv^(2)`……..ii
As per question, maximum speed of photoelectron in two cases differ by a factor 2
From eqn (i) and (ii)
`implies((hc)/(lamda_(1))-phi)/((hc)/(lamda_(2))-phi)=4implies(hc)/(lamda_(1))-phi=(4hc)/(lamda_(2))-4phi`
`implies(4hc)/(lamda_(2))-(hc)/(lamda_(1))=3 phi implies phi = 1/3 hc(4/(lamda_(2))-1/(lamda_(1)))`
`=1/3xx1240((4xx350-540)/(350xx540))=1.8eV`