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(D) The energy of the incident photons is $E = 10 eV$.The threshold frequency is $
u_0 = 2 	imes 10^{15} Hz$.The work function $W_0$ is given by $W_

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$1.7$

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(D) The energy of the incident photons is $E = 10 eV$.The threshold frequency is $
u_0 = 2 	imes 10^{15} Hz$.The work function $W_0$ is given by $W_0 = h
u_0$.Substituting the values: $W_0 = (6.63 	imes 10^{-34} Js) 	imes (2 	imes 10^{15} Hz) = 13.26 	imes 10^{-19} J$.To convert this into $eV$,divide by the charge of an electron $(1.6 	imes 10^{-19} C)$:$W_0 = \frac{13.26 	imes 10^{-19}}{1.6 	imes 10^{-19}} eV \approx 8.2875 eV \approx 8.3 eV$.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - W_0$.$K_{max} = 10 eV - 8.3 eV = 1.7 eV$.

Photons of energy $10 eV$ are incident on a photosensitive surface of threshold frequency $2 \times 10^{15} Hz$. The kinetic energy in $eV$ of the photoelectrons emitted is [Planck's constant $h = 6.63 \times 10^{-34} Js$]. (in $eV$)

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