Question
In a photoelectric effect experiment, it is seen that when light of wavelength is incident on a metal, the measured photoelectric current drops to zero at a potential
. Calculate the work function of the metal and its cut-off frequency for photoelectric effect.
Answer :
Word Count : 185
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The energy of incident photons is given by [ E = \frac{hc}{\lambda} ] Here, (h = 6.626 \times 10^{-34},\mathrm{Js}), (c = 3.00 \times 10^8,\mathrm{m/s}), and (\lambda = 180,\mathrm{nm} = 180 \times 10^{-9},\mathrm{m}). [ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{180 \times 10^{-9}} ] [ E = \frac{19.878 \times 10^{-26}}{180 \times 10^{-9}} = \frac{19.878 \times 10^{-26}}{1.8 \times 10^{-7}} ] [
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