# 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? - Physics

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?

#### Solution

Work function of caesium metal, phi_o = 2.14eV

Frequency of light, v = 6.0 xx 10^14 Hz

(a)The maximum kinetic energy is given by the photoelectric effect as:

K = hv - phi_o

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 V_0, we can write the equation for kinetic energy as:

K =eV_o

:. V_o = 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 xx 10^12

:. v = 3.323  xx 10^5 "m/s" = 332.3 "km/s"

Hence, the maximum speed of the emitted photoelectrons is  332.3 km/s.

Concept: Electron Emission
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#### APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 11 Dual Nature of Radiation and Matter
Q 2 | Page 407