1.

When frequencies f_(1) and f_(2) are incident on two indentical photo sensitive surfaes ,maximum velocities of photo-electrons of mass m are v_(1) and v_(2) hence……

Answer»

`v_(1)^(2)-v_(2)^(2)=(2h)/(m)(f_(1)-f_(2))`
`v_(1)+v_(2)=[(2h)/(m)(f_(1)+f_(2))]^((1)/(2))`
`v_(1)^(2)+v_(2)^(2)=(2h)/(m)(f_(1)+f_(2))`
`v_(1)-v_(2)=[(2h)/(m)(f_(1)+f_(2))]^((1)/(2))`

Solution :From Eiestien.s PHOTOELECTRIC equation,
`(1)/(2)mv_(max)^(2)=HF-phi`
`therefore hf=(1)/(2)mv_(max)^(2)+phi`
`therefore hf_(1)=(1)/(2)mv_(1)^(2)+phi [because v_(max)=v_(1)]`
and `hf_(2)=(1)/(2)mv_(2)^(2)+phi[because v_(max)=v_(2)]`
`therefore H(f_(1)-f_(2))=(1)/(2)m(v_(1)^(2)-v_(2)^(2))`
`therefore (2h)/(m)(f_(1)-f_(2))=v_(1)^(2)therefore v_(1)^(2)-v_(2)^(2)=(2h)/(m)=(2h)/(m)(f_(1)-f_(2))`


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