Consider a metal exposed to light of wavelength 600nm. The maximum energy of the electrons doubles when light of wavelength 400nm is used. Find the work function in eV.

Explanation

Here, `lambda=600nm, K'_(max)=2K_(max), lambda'=400nm, phi_(0)=?`
`K_(max)=(hc)/lambda-phi_(0)......(i)`
`2K_(max)=(hc)/(lambda')-phi_(0)......(ii)`
Dividing (ii) by (i), we get `2=((hc//lambda')-phi_(0))/((hc//lambda)-phi_(0)) or (2hc)/lambda-2phi_(0)=(hc)/(lambda')-phi_(0) or hc(2/lambda-1/(lambda'))=phi_(0)`
`:. phi_(0)=1240(2/600-1/400)=1.03eV [:' hc~=1240 eVnm]`