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Consider a metal exposed to light of wav...

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.

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Give , wavelength of light used initially ,
`lamda= 600 nm `
Let `K_(max)` be the maximum energy of electron for `lamda_1` .
For `lamda.= 400 nm,`
`K._(max)=2K_(max)`
Also, `K_(max)=hv-phi_(0)=((hc)/(lamda))-phi_0" "....(i)`
and `K._(max)=2K_(max)=((hc)/(lamda.))-phi_0" "...(ii)`
where `phi_0` is the work function of metal
Dividing (ii) by (i) , we have
`2=((hc//lamda.)-phi_0)/((hc//lamda)-phi_0)`
`implies(2hc)/lamda-2phi_0=(hc)/(lamda.)-phi_0`
`implieshc(2/lamda-1/(lamda.))=phi_0`
Then `phi_0=1240(2/600-1/400)=eV=1.03eV`
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