1.

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)

Answer»

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

SOLUTION :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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