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(D) According to Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - W_0 = \frac{1}{2}mv^2$,where $W_0 = \frac{hc}{\lambda_0}$ is the

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

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(D) According to Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - W_0 = \frac{1}{2}mv^2$,where $W_0 = \frac{hc}{\lambda_0}$ is the work function.Thus,$v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right)} \dots (i)$When the wavelength is changed to $\lambda' = \frac{3\lambda}{4}$,the new velocity $v'$ is:$v' = \sqrt{\frac{2hc}{m} \left( \frac{1}{3\lambda/4} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \left( \frac{4\lambda_0 - 3\lambda}{3\lambda \lambda_0} \right)} \dots (ii)$Dividing $(ii)$ by $(i)$:$\frac{v'}{v} = \sqrt{\frac{4\lambda_0 - 3\lambda}{3\lambda \lambda_0} \cdot \frac{\lambda \lambda_0}{\lambda_0 - \lambda}} = \sqrt{\frac{4}{3} \cdot \frac{\lambda_0 - 0.75\lambda}{\lambda_0 - \lambda}}$Since $\lambda_0 > \lambda$,it follows that $(\lambda_0 - 0.75\lambda) > (\lambda_0 - \lambda)$.Therefore,$\frac{\lambda_0 - 0.75\lambda}{\lambda_0 - \lambda} > 1$.This implies $\frac{v'}{v} > \sqrt{\frac{4}{3}}$,or $v' > v(4/3)^{1/2}$.

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

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