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When a metallic surface is illuminated b...

When a metallic surface is illuminated by a light of frequency `8xx10^(14)`Hz, photoelectron of maximum energy 0.5 eV is emitted. When the same surface is illuminated by light of frequency `12xx10^(14)` Hz, photoelectron of maximum energy 2 eV is emitted. The work function is

Text Solution

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The correct Answer is:
`2.5`
`8xx10^(14) h = phi_(0)+0.5 " "…(i)`
`12 xx 10^(14) h = phi_(0)+2 " "…(ii)`
From eq .(i) & (ii) , we get
`12/8 = (phi_(0)+2)/(phi_(0)+0.5)" "rArr3/2 =(phi_(0)+2)/(phi_(0)+0.5)`
`rArr 3phi_(0)+1.5 = 2phi_(0) +4`
`rArr phi_(0)=2.5` eV
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