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(B) The work function $\phi_0$ of a photoelectric emitter is given by the formula $\phi_0 = \frac{hc}{\lambda_0}$,where $h$ is Planck's constant,$c$ i

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$2: 1$

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(B) The work function $\phi_0$ of a photoelectric emitter is given by the formula $\phi_0 = \frac{hc}{\lambda_0}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda_0$ is the threshold wavelength.From this relation,we can see that $\phi_0 \propto \frac{1}{\lambda_0}$.For the first emitter,the incident wavelength is $300 \ nm$. Since photoelectrons are emitted,the threshold wavelength $\lambda_{0_1}$ must be at least $300 \ nm$.For the second emitter,the threshold wavelength $\lambda_{0_2}$ is given as $600 \ nm$.Assuming the first emitter is just at the threshold for $300 \ nm$,we have $\lambda_{0_1} = 300 \ nm$ and $\lambda_{0_2} = 600 \ nm$.The ratio of the work functions is $\frac{\phi_{0_1}}{\phi_{0_2}} = \frac{\lambda_{0_2}}{\lambda_{0_1}} = \frac{600 \ nm}{300 \ nm} = \frac{2}{1}$.Thus,the ratio is $2: 1$.

When a light of wavelength $300 \ nm$ falls on a photoelectric emitter,photoelectrons are emitted. 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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