1.

A stopping potential of 0.82 volt is required to stop the photoelectrons emitted from a metallic surface by light of wavelength 4000 Å. The stopping potential for wavelength 3000 Å will be :

Answer»

1.85 V
2.85 V
3.0 V
4.1 V.

Solution :`0.82eV=(12375)/(4000)-phi_(0)`
`:. Phi_(0)=2*27eV`
Also `eV_(S)=(12375)/(3000)-2.27`
Then `V_(S)=1*85V`.


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