11.8 The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.

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    Payal Gupta | Contributor-Level 10

    4 months ago

    11.8 Threshold frequency of the metal, ν0=3.3×1014 Hz

    Frequency of the light incident on metal, ν=8.2×1014 Hz

    Charge of an electron, e = 1.6 ×10-19 C

    Planck’s constant, h = 6.626 ×10-34 Js

    Let the cut-off voltage for the photoelectric emission from the metal be V0

    The equation of the cut-off energy is given as:

    eV0 = h(ν-ν0) or

    V0=h(ν-ν0)e = 6.626×10-34×(8.2×1014-3.3×1014)1.6×10-19 V = 2.0292 V

    Therefore, the cut-off voltage for the photoelectric emission is 2.0292V

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