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(C) According to Einstein's photoelectric equation,the kinetic energy $K$ is given by: $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work functi

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$\lambda > \lambda' > \lambda/2$

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(C) According to Einstein's photoelectric equation,the kinetic energy $K$ is given by: $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For the initial state: $K = \frac{hc}{\lambda} - \phi$  --- $(1)$For the final state,where kinetic energy is $2K$: $2K = \frac{hc}{\lambda'} - \phi$ --- $(2)$From $(1)$,we have $\frac{hc}{\lambda} = K + \phi$. Since $K > 0$,it follows that $\frac{hc}{\lambda} > \phi$.From $(2)$,$\frac{hc}{\lambda'} = 2K + \phi$. Since $K > 0$,it follows that $\frac{hc}{\lambda'} > \phi$.Dividing $(1)$ by $(2)$: $\frac{\lambda'}{\lambda} = \frac{K + \phi}{2K + \phi}$.Since $K + \phi Also,$\frac{\lambda'}{\lambda} = \frac{K + \phi}{2K + \phi} = \frac{1}{2} \left( \frac{2K + 2\phi}{2K + \phi} \right) = \frac{1}{2} \left( \frac{2K + \phi + \phi}{2K + \phi} \right) = \frac{1}{2} \left( 1 + \frac{\phi}{2K + \phi} \right)$.Since $\frac{\phi}{2K + \phi} > 0$,the term in the parenthesis is greater than $1$. Therefore,$\frac{\lambda'}{\lambda} > \frac{1}{2}$,which implies $\lambda' > \lambda/2$.Combining these,we get $\lambda > \lambda' > \lambda/2$.

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K$. If $K$ is to be increased to twice its original value,the wavelength must be changed to $\lambda'$ such that :-

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