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Question

(B) The work function $\phi_0$ of a photoelectric emitter is related to the threshold wavelength $\lambda_0$ by the equation $\phi_0 = \frac{hc}{\lamb

Vedclass · Accepted Answer

$2 : 1$

Vedclass · Answer

(B) The work function $\phi_0$ of a photoelectric emitter is related to the threshold wavelength $\lambda_0$ by the equation $\phi_0 = \frac{hc}{\lambda_0}$.Given that the photoelectrons are just liberated,the incident wavelength is equal to the threshold wavelength.For the first emitter,$\lambda_{01} = 300 \ nm$.For the second emitter,$\lambda_{02} = 600 \ nm$.The ratio of the work functions is $\frac{\phi_{01}}{\phi_{02}} = \frac{hc / \lambda_{01}}{hc / \lambda_{02}} = \frac{\lambda_{02}}{\lambda_{01}}$.Substituting the values: $\frac{\phi_{01}}{\phi_{02}} = \frac{600 \ nm}{300 \ nm} = \frac{2}{1}$.Thus,the ratio is $2 : 1$.

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

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