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(C) The Einstein's photoelectric equation is given by $K_{max} = \frac{hc}{\lambda} - \phi$.Initially,$5\,eV = \frac{hc}{\lambda} - \phi$  ..........$

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$3$

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(C) The Einstein's photoelectric equation is given by $K_{max} = \frac{hc}{\lambda} - \phi$.Initially,$5\,eV = \frac{hc}{\lambda} - \phi$  ..........$(1)$When the wavelength is decreased by $60\%$,the new wavelength $\lambda' = \lambda - 0.6\lambda = 0.4\lambda = \frac{2}{5}\lambda$.Substituting this into the equation for the new kinetic energy:$17\,eV = \frac{hc}{0.4\lambda} - \phi = \frac{2.5hc}{\lambda} - \phi$  ..........$(2)$From equation $(1)$,we have $\frac{hc}{\lambda} = 5\,eV + \phi$.Substitute this into equation $(2)$:$17\,eV = 2.5(5\,eV + \phi) - \phi$$17\,eV = 12.5\,eV + 2.5\phi - \phi$$17\,eV - 12.5\,eV = 1.5\phi$$4.5\,eV = 1.5\phi$$\phi = \frac{4.5}{1.5}\,eV = 3\,eV$.Thus,the work function of the metal is $3\,eV$.

If in an experiment of photoelectric effect,the wavelength of light is decreased by $60\%$,then $K_{max}$ changes from $5\,eV$ to $17\,eV$. The work function of the metal is ............. $eV$.

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