The threshold frequency for a certain metal is 3.3 × 1014 Hz. If lightof frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
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Solution
Given: The threshold frequency of the metal 3.3×1014Hz, the frequency of light incident on the metal is 8.2×1014Hz.
The cutoff voltage is given as,
V0=h(ν−ν0)e
Where, the threshold frequency of metal is ν0, the frequency of the light is ν, Planck’s constant is h, the cut off voltage is V0 and the charge on an electron is e.
By substituting the given values in the above equation, we get V0=6.626×10−34(8.2×1014−3.3×1014)1.6×10−19=6.626×10−34(4.9×1014)1.6×10−19=2.03V
Thus, the cutoff voltage for the photoelectric emission is 2.03V.