The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
Work function of caesium metal, ø0 = 2.14 eV
Frequency of light, v = 6.0 x 1014 Hz
(a) The maximum kinetic energy is given by the photoelectric effect as: K = hv - ø0
Where,
h = Planck’s constant = 6.626 × 10−34 Js
∴ K = 6.626 x 1034 x 6 x 1014 / 1.6 x 10-19 - 2.14 = 2.485 - 2.14 = 0.345 eV
Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.
( b ) For stopping potential , V0 we can write the equation for kinetic energy as: K = eV0
∴ V0 = K/e = 0.345 x 1.6 x 10-19 / 1.6 x 10-19 = 0.345 V
Hence, the stopping potential of the material is 0.345 V.
(c) Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as: K = 1/2 mv2
Where,
m = Mass of an electron = 9.1 × 10−31 kg
v2 = 2K/m = 2 x 0.345 x 1.6 x 10-19 / 9.1 x 10-31 = 0.1104 x 1012
∴ v = 3.323 x 105 m/s = 332.3 km/s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.
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What is the energy of incident photons?
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