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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)_{max}$ is given by: $(K.E.)_{max} = \frac{hc}{\lambda} - W_0$,wh

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greater than $(1.5)^{1/2} V$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)_{max}$ is given by: $(K.E.)_{max} = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.For the first case: $(K.E.)_1 = \frac{1}{2}mV^2 = \frac{hc}{\lambda} - W_0$ ... $(1)$For the second case,with wavelength $\lambda' = \frac{2\lambda}{3}$:$(K.E.)_2 = \frac{1}{2}mv_2^2 = \frac{hc}{\lambda'} - W_0 = \frac{hc}{(2\lambda/3)} - W_0 = \frac{3}{2}\frac{hc}{\lambda} - W_0$ ... $(2)$From equation $(1)$,$\frac{hc}{\lambda} = \frac{1}{2}mV^2 + W_0$. Substituting this into equation $(2)$:$(K.E.)_2 = \frac{3}{2}(\frac{1}{2}mV^2 + W_0) - W_0$$(K.E.)_2 = \frac{3}{2}(\frac{1}{2}mV^2) + \frac{3}{2}W_0 - W_0$$(K.E.)_2 = \frac{3}{2}(\frac{1}{2}mV^2) + \frac{1}{2}W_0$Since $W_0 > 0$,it follows that $(K.E.)_2 > \frac{3}{2}(K.E.)_1$.$\frac{1}{2}mv_2^2 > \frac{3}{2}(\frac{1}{2}mV^2)$$v_2^2 > 1.5 V^2$$v_2 > \sqrt{1.5} V = (1.5)^{1/2} V$.

When light of wavelength $\lambda$ falls on the emitter of a photocell,the maximum speed of the emitted photoelectrons is $V$. If the incident wavelength is changed to $\frac{2\lambda}{3}$,the maximum speed of the emitted photoelectrons will be:

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