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(D) According to Einstein's photoelectric equation: $\frac{1}{2}m\upsilon^2 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the wo

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$ > \,\,\upsilon \,{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$

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(D) According to Einstein's photoelectric equation: $\frac{1}{2}m\upsilon^2 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.Thus,$\upsilon = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} ight)} = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} ight)} \dots (1)$When the wavelength is changed to $\lambda' = \frac{3\lambda}{4}$,the new velocity $\upsilon'$ is:$\upsilon' = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - 3\lambda/4}{(3\lambda/4) \lambda_0} ight)} \dots (2)$Dividing equation $(2)$ by $(1)$:$\frac{\upsilon'}{\upsilon} = \sqrt{\frac{\lambda_0 - 3\lambda/4}{3\lambda/4 \lambda_0} 	imes \frac{\lambda \lambda_0}{\lambda_0 - \lambda}} = \sqrt{\frac{4}{3} \left( \frac{\lambda_0 - 3\lambda/4}{\lambda_0 - \lambda} ight)}$Since $\lambda_0 - 3\lambda/4 > \lambda_0 - \lambda$,the term $\frac{\lambda_0 - 3\lambda/4}{\lambda_0 - \lambda} > 1$.Therefore,$\upsilon' > \upsilon \left( \frac{4}{3} ight)^{1/2}$.

When radiation of wavelength $\lambda$ is incident on a photoelectric cell,the maximum velocity of the emitted photoelectrons is $\upsilon$. If the wavelength of the incident radiation is changed to $3\lambda / 4$,the maximum velocity of the emitted photoelectrons will be ..........

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