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(A) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$,where $\phi$ is the work function.For wavelength $\l

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$> v \left( \frac{4}{3} \right)^{1/2}$

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(A) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$,where $\phi$ is the work function.For wavelength $\lambda$: $\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$  --- $(1)$For wavelength $\lambda' = \frac{3\lambda}{4}$: $\frac{hc}{\lambda'} = \frac{4hc}{3\lambda} = \phi + \frac{1}{2}mv'^2$ --- $(2)$From $(1)$,$\frac{hc}{\lambda} = \phi + K$,where $K = \frac{1}{2}mv^2$. Thus,$\frac{hc}{3\lambda} = \frac{\phi + K}{3}$.Substituting into $(2)$: $\frac{4}{3}(\phi + K) = \phi + \frac{1}{2}mv'^2$.$\frac{1}{2}mv'^2 = \frac{4}{3}\phi + \frac{4}{3}K - \phi = \frac{1}{3}\phi + \frac{4}{3}K$.Since $\phi > 0$,$\frac{1}{2}mv'^2 > \frac{4}{3}K$.$\frac{1}{2}mv'^2 > \frac{4}{3} \left( \frac{1}{2}mv^2 \right) \implies v'^2 > \frac{4}{3}v^2$.Therefore,$v' > v \left( \frac{4}{3} \right)^{1/2}$.

Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be:

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