11.26 Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity ( 105 W m–2) red light of wavelength 6328 Å produced by a He-Ne laser?

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    11.26 Wavelength of ultraviolet light, λ = 2271Å = 2271 ×10-10 m

    Stopping potential of the metal, V0 = 1.3 V

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

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

    Speed of light, c = 3 ×108 m/s

    Work function of the metal, 0

    Frequency of light = ν

    We have the photo-energy relation from the photoelectric effect as:

    0=hν-eV0

    hcλ-eV0

    6.626×10-34×3×1082271×10-10-1.6×10-19×1.3

    = 8.75 ×10-19 - 2.08 ×10-19

    = 6.67 ×10-19 J

    6.67×10-191.6×10-19 eV

    = 4.17 eV

    Let ν0 be the threshold frequency of the metal.

    Therefore, 0 = h ν0

    ν0= 0h = 6.67×10-196.626×10-34 Hz = 1.007 ×1015 Hz

    Wavelength of red light, λr&nbs

    ...more

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