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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:$K_{\max} = hf - \phi_0$Where $h$ is Planck's con

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:$K_{\max} = hf - \phi_0$Where $h$ is Planck's constant,$f$ is the frequency of incident radiation,and $\phi_0$ is the work function.First,calculate the energy of the incident photon in Joules:$E = hf = (6.63 	imes 10^{-34} \ J \cdot s) 	imes (6 	imes 10^{14} \ Hz) = 39.78 	imes 10^{-20} \ J$Convert this energy into electron-volts $(eV)$:$E_{eV} = \frac{39.78 	imes 10^{-20} \ J}{1.6 	imes 10^{-19} \ J/eV} = 24.8625 	imes 10^{-1} \ eV = 2.48625 \ eV \approx 2.49 \ eV$Now,calculate the maximum kinetic energy:$K_{\max} = 2.49 \ eV - 2 \ eV = 0.49 \ eV$.

When radiation of frequency $6 \times 10^{14} \ Hz$ is incident on a metal surface with a work function of $2 \ eV$,the maximum kinetic energy of the emitted photoelectrons is ........ $eV$ (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$,$1 \ eV = 1.6 \times 10^{-19} \ J$). (in $.49$)

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