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(A) The maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by Einstein's photoelectric equation:$K_{max} = E - \Phi$where $E$ is th

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(A) The maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by Einstein's photoelectric equation:$K_{max} = E - \Phi$where $E$ is the energy of the incident photon and $\Phi$ is the work function of the metal.The work function $\Phi$ is given by $\Phi = h
u_0$,where $
u_0$ is the threshold frequency.Given $h = 6 	imes 10^{-34} \ J \cdot s$ and $
u_0 = 1.6 	imes 10^{15} \ Hz$:$\Phi = (6 	imes 10^{-34} \ J \cdot s) 	imes (1.6 	imes 10^{15} \ Hz) = 9.6 	imes 10^{-19} \ J$.To convert the work function into $eV$,divide by $1.6 	imes 10^{-19} \ J/eV$:$\Phi = \frac{9.6 	imes 10^{-19}}{1.6 	imes 10^{-19}} \ eV = 6 \ eV$.Now,calculate the maximum kinetic energy:$K_{max} = 8 \ eV - 6 \ eV = 2 \ eV$.

$A$ photon of energy $8 \ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \ Hz$. The maximum kinetic energy of the photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)

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