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Calculate the work function of the metal...

Calculate the work function of the metal, if the kinetic energies of the photoelectrons are `E_(1)` and `E_(2)`, with wavelengths of incident light `lambda_(1)` and `lambda_(2)`

A

`(E_(1)lambda_(1)-E_(2)lambda_(2))/(lambda_(2)-lambda_(1))`

B

`(E_(1)E_(2))/(lambda_(1)-lambda_(2))`

C

`((E_(1)-E_(2))lambda_(1)lambda_(2))/((lambda_(1)-lambda_(2)))`

D

`(lambda_(1)lambda_(2)E_(1))/((lambda_(1)-lambda_(2))E_(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
We know that `E_(1)=(hc)/(lambda_(1))-W" "....(i)`
`E_(2)=(hc)/(lambda_(2))-W" "......(ii)`
From Eqs. (i) and (ii), we get
`therefore" "(E_(1)+W)/(E_(2)+W)=(lambda_(2))/(lambda_(1))" or "W=(E_(1)lambda_(1)-E_(2)lambda_(2))/((lambda_(2)-lambda_(1)))`
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