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

Explanation

Showing maximum kinetic energy `(1)/(2) mv_(max)^(2)` by symbol K,according to photoelectric equation ,
`hv=hv_(0)+k`
`therefore (hc)/(lambda)=phi_(0)+k`
(i)For `lambda_(1),(hc)/(lamba_(1))=phi_(0)+K_(1)` …….(1)
(ii) For `lambda_(2),(hc)/(lambda_(2))=phi_(0)+2K_(1)` ......(2)
`(because K_(2)=2K_(1))`
From equation (1) and (2)
`(hc)/(lambda_(2))=phi_(0)+2((hc)/(lambda_(1))-phi_(0))`
`therefore phi_(0)=hc((2)/(lambda_(1))-(1)/(lambda_(2)))`
`therefore phi_(0)=6.625x10^(-34)xx3xx10^(8)`
`xx((2)/(600xx10^(-9))-(1)/(400xx10^(-9)))`
`therefore =(6.625xx10^(-34)xx3xx10^(8))((2)/(600)-(1)/(400))`
`therefore phi_(0)=(6.625xx10^(-34)xx3xx10^(8))/(10^(-9))((200)/(600xx400))`
`=(6.625xx10^(-34)xx3xx10^(8))/(10^(-9))xx(1)/(1200)`
`=1.656xx10^(-9)J`
`therefore phi_(0)=(1.656xx10^(-19))/(1.6xx10^(-19))eV=1.035 eV`