When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

Options

A. `1:2`
B. `2:1`
C. `4:1`
D. `1:4`

Answer

B

Explanation

Work function `phi=hc//lambda`
`therefore phi prop (1)/(lambda` or, `(phi_(1))/(phi_(2))=(lambda_(2))/(lambda_(1))=(600)/(300)=(2)/(1)`