1.

When a metallic surface is illuminated by light of frequency 8xx10^(14)Hz a photoelectron of energy 0.5eV 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 :

Answer»

0.5 eV
1.5 eV
2.5 eV
3.5 eV

Solution :`8xx10^(14)H= phi_(0)+0*5`
`12xx10^(14)h=phi_(0)+2`
Then `(12)/(8)=(phi_(0)+2)/(phi_(0)+0*5)`
SOLVING `phi_(0)=2*5eV`


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