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(C) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = K_{\max} + \phi$.Given that the work function $\phi$ is negligible,we have $

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$\lambda=\left(\frac{2 mc}{h}\right) \lambda_{d}^{2}$

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(C) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = K_{\max} + \phi$.Given that the work function $\phi$ is negligible,we have $\frac{hc}{\lambda} = K_{\max}$.The de-Broglie wavelength of the emitted photoelectron is given by $\lambda_{d} = \frac{h}{\sqrt{2mK_{\max}}}$.Squaring both sides,we get $\lambda_{d}^{2} = \frac{h^{2}}{2mK_{\max}}$,which implies $K_{\max} = \frac{h^{2}}{2m\lambda_{d}^{2}}$.Substituting the expression for $K_{\max}$ into the photoelectric equation: $\frac{hc}{\lambda} = \frac{h^{2}}{2m\lambda_{d}^{2}}$.Solving for $\lambda$,we get $\lambda = \frac{hc \cdot 2m\lambda_{d}^{2}}{h^{2}} = \left(\frac{2mc}{h}\right) \lambda_{d}^{2}$.

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectron emitted from the surface has mass $m$ and de-Broglie wavelength $\lambda_{d}$,then

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