Question

If light of wavelength `lambda_(1)` is allowed to fall on a metal , then kinetic energy of photoelectrons emitted is `E_(1)`. If wavelength of light changes to `lambda_(2)` then kinetic energy of electrons changes to `E_(2)`. Then work function of the metal is
A. `(E_(1) E_(2) (lambda_(1) - lambda_(2)))/(lambda_(1) lambda_(2))`
B. `(E_(1) lambda_(1) - E_(2) lambda_(2))/((lambda_(1) - lambda_(2))`
C. `(E_(1) lambda_(1) - E_(2) lambda_(2))/((lambda_(2) - lambda_(1)))`
D. `(lambda_(1) lambda_(2) E_(1) E_(2))/((lambda_(2) - lambda_(1)))`

Answer 1

Correct Answer - C
`E = W_(0) + K_(max) rArr (hc)/(lambda_(1)) = W_(0) + E_(1)` and `(hc)/(lambda_(2)) = W_(0) + E_(2)`
`rArr hc = W_(0) lambda_(1) + E_(1) lambda_(1)` and `hc = W_(0) lambda_(2) + E_(2) lambda_(2)`
`rArr W_(0) lambda_(1) + E_(1) lambda_(1) = W_(0) lambda_(2) + E_(2) lambda_(2) rArr W_(0) = (E_(1) lambda_(1) - E_(2) lambda_(2))/((lambda_(2) - lambda_(1)))`.