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(C) According to Einstein's photoelectric equation,the kinetic energy $E$ of the ejected electrons is given by: $E = \frac{hc}{\lambda} - \Phi$,where

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$\lambda_1 > \lambda / 2$

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(C) According to Einstein's photoelectric equation,the kinetic energy $E$ of the ejected electrons is given by: $E = \frac{hc}{\lambda} - \Phi$,where $\Phi$ is the work function of the surface.For the initial state: $E = \frac{hc}{\lambda} - \Phi$.For the final state,the energy becomes $2E$: $2E = \frac{hc}{\lambda_1} - \Phi$.From the first equation,$\Phi = \frac{hc}{\lambda} - E$.Substituting this into the second equation: $2E = \frac{hc}{\lambda_1} - (\frac{hc}{\lambda} - E)$.$2E = \frac{hc}{\lambda_1} - \frac{hc}{\lambda} + E$.$E = \frac{hc}{\lambda_1} - \frac{hc}{\lambda}$.Since $E > 0$,it follows that $\frac{hc}{\lambda_1} > \frac{hc}{\lambda}$,which implies $\lambda_1 Also,$\frac{hc}{\lambda_1} = E + \frac{hc}{\lambda}$.Since $E = \frac{hc}{\lambda} - \Phi$,we have $\frac{hc}{\lambda_1} = \frac{hc}{\lambda} - \Phi + \frac{hc}{\lambda} = \frac{2hc}{\lambda} - \Phi$.Since $\Phi > 0$,$\frac{hc}{\lambda_1} \frac{\lambda}{2}$.Thus,$\frac{\lambda}{2} < \lambda_1 < \lambda$.

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with energy $E$. If $E$ is to be increased to twice the original value,the wavelength changes to $\lambda_1$. Which of the following is correct?

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