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(D) According to Einstein's photoelectric equation: $E = W_0 + K_{\max}$,where $K_{\max} = \frac{1}{2}mv_{\max}^2$.Thus,$\frac{1}{2}mv^2 = \frac{hc}{\

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$> v(4/3)^{1/2}$

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(D) According to Einstein's photoelectric equation: $E = W_0 + K_{\max}$,where $K_{\max} = \frac{1}{2}mv_{\max}^2$.Thus,$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - W_0$,which implies $v = \sqrt{\frac{2}{m} (\frac{hc}{\lambda} - W_0)}$.For the new wavelength $\lambda' = \frac{3\lambda}{4}$,the new energy is $E' = \frac{hc}{\lambda'} = \frac{hc}{3\lambda/4} = \frac{4}{3}E$.The new maximum velocity $v'$ is given by $v' = \sqrt{\frac{2}{m} (\frac{4}{3}E - W_0)}$.We can rewrite this as $v' = \sqrt{\frac{4}{3} \cdot \frac{2}{m} (E - \frac{3}{4}W_0)}$.Since $W_0 > \frac{3}{4}W_0$,it follows that $(E - \frac{3}{4}W_0) > (E - W_0)$.Therefore,$v' > \sqrt{\frac{4}{3}} \cdot \sqrt{\frac{2}{m}(E - W_0)}$,which simplifies to $v' > v(4/3)^{1/2}$.

When radiation of wavelength $\lambda$ is incident on a photocell,the maximum velocity of the photoelectrons is $v$. What will be the maximum velocity of the photoelectrons when radiation of wavelength $3\lambda/4$ is incident on the photocell?

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