Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $\lambda_{1}=360 \mathrm{~nm}$ and $\lambda_{2}=600 \mathrm{~nm} .$ What is the ratio of work functions for lights of wavelength ${ }^{\prime} \lambda_{1}{ }^{\prime}$ to $\lambda_{2}{ }^{\prime}$?

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Aicharya · Accepted Answer

$\frac{h c}{\lambda_{1}}-\phi_{1}=\mathrm{KE} \quad \quad \lambda_{1}=360 \mathrm{~nm}=360 	imes 10^{-9} \mathrm{~m}$ $\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{2}=\mathrm{KE} \quad \quad\lambda_{2}=600 \mathrm{~nm}=600 	imes 10^{-9} \mathrm{~m}$ $\frac{h c}{\lambda_{1}}=\phi_{1} \quad \quad \phi \propto \frac{1}{\lambda}$ $	herefore \frac{\phi_{1}}{\phi_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{600}{360}=\frac{10}{6}=\frac{5}{3}$

Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $\lambda_{1}=360…

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