If the sodium metal is irradiated with a wavelength 450 nm, calculate the kinetic energy and the velocity of the ejected photoelectron. ( Given work function = 2.3 eV).

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Dear student,E=W+KE,,So E=hc/lamba=6.626×10^-34×3×10^8/450×10^-9 =0.04417×10^-17 joules...Now Kinetic energy=4.42×10^-19-3.68×10^-19=7.4×10^-20 joules..We know Kinetic energy=1/2mv^2=7.4×10^-20,,so v^2=1.626×10^11=4.03×10^5m/s