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Question
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
- maximum kinetic energy of the emitted electrons,
- Stopping potential, and
- maximum speed of the emitted photoelectrons?
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Solution
Work function of caesium metal, `phi_0` = 2.14 eV
Frequency of light, v = 6.0 × 1014 Hz
(a) The maximum kinetic energy is given by the photoelectric effect as:
K = hv − `phi_0`
Where,
h = Planck’s constant = 6.626 × 10−34 Js
∴ K = `(6.626 xx 10^-34 xx 6 xx 10^14)/(1.6 xx 10^(-19)) - 2.14`
= 2.485 − 2.140
= 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
∴ `V_0 = K/e`
= `(0.345 xx 1.6 xx 10^(-19))/(1.6 xx 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 mv^2`
Where,
m = Mass of an electron = 9.1 × 10−31 kg
`v^2 = (2K)/m`
= `(2 xx 0.345 xx 1.6 xx 10^(-19))/(9.1 xx 10^(-31))`
= 0.1104 × 1012
∴ v = 3.323 × 105 m/s
= 332.3 km/s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.
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