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(D) According to Einstein's photoelectric equation: $K_{max} = \frac{1}{2} m u^2 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is t

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greater than $(4/3)^{1/2} u$

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(D) According to Einstein's photoelectric equation: $K_{max} = \frac{1}{2} m u^2 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.For wavelength $\lambda$: $\frac{1}{2} m u^2 = \frac{hc}{\lambda} - \phi$  ---$(1)$For wavelength $\lambda' = \frac{3\lambda}{4}$: $\frac{1}{2} m u_1^2 = \frac{hc}{\lambda'} - \phi = \frac{4hc}{3\lambda} - \phi$ ---$(2)$From $(1)$,$\frac{hc}{\lambda} = \frac{1}{2} m u^2 + \phi$.Substitute this into $(2)$:$\frac{1}{2} m u_1^2 = \frac{4}{3} (\frac{1}{2} m u^2 + \phi) - \phi$$\frac{1}{2} m u_1^2 = \frac{4}{3} (\frac{1}{2} m u^2) + \frac{4}{3} \phi - \phi$$\frac{1}{2} m u_1^2 = \frac{4}{3} (\frac{1}{2} m u^2) + \frac{1}{3} \phi$Since $\phi > 0$,it follows that $\frac{1}{2} m u_1^2 > \frac{4}{3} (\frac{1}{2} m u^2)$.$u_1^2 > \frac{4}{3} u^2 \Rightarrow u_1 > \sqrt{\frac{4}{3}} u$.

In a photocell,the incident wavelength is $\lambda$. The maximum speed of the emitted photoelectrons is $u$. If the incident wavelength is changed to $3\lambda / 4$,then the maximum speed of the emitted photoelectrons will be:

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