In a photoemissive cell with exciting wavelen | vedclass

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{h

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

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.Thus,$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = hc \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right)$.So,$v = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right)}$.When the wavelength is changed to $\lambda' = \frac{3\lambda}{4}$,the new speed $v'$ is:$v' = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - 3\lambda/4}{(3\lambda/4) \lambda_0} \right)}$.Taking the ratio:$\frac{v'}{v} = \sqrt{\frac{\lambda_0 - 3\lambda/4}{3\lambda/4 \lambda_0} \cdot \frac{\lambda \lambda_0}{\lambda_0 - \lambda}} = \sqrt{\frac{4}{3}} \sqrt{\frac{\lambda_0 - 3\lambda/4}{\lambda_0 - \lambda}}$.Since $\lambda_0 > \lambda$,it follows that $\lambda_0 - 3\lambda/4 > \lambda_0 - \lambda$.Therefore,$\sqrt{\frac{\lambda_0 - 3\lambda/4}{\lambda_0 - \lambda}} > 1$.This implies $v' > \sqrt{\frac{4}{3}} v$.

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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