Maximum Kinetic Energy of Photon and Stopping Potential

Example – 01:

  • When a radiation of certain wavelength shines on the cathode of the photoelectric cell, the photocurrent produced can be reduced to zero by applying stopping potential of 2.63 V. If the work function of the photoemitter is 4 eV, find the wavelengths of radiation.
  • Given: Stopping potential = Vs = 2.63 V, work function = Φ = 4 eV = 4 x 1.6 x 10-19 J = 6.4 x 10-19 J, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
  • To Find: Wavelength of radiation = λ = ?
  • Solution:

By Einstein’s photoelectric equation

K.E. max = hν – Φ  = hc/λ – Φ

But K.E. max = eVs

 eVs =  hc/λ – Φ



∴ (1.6 x 10-19) x(2.63) = (6.63 x 10-34)(3 x 108)/λ – 6.4 x 10-19

∴ 4.21 x 10-19 + 6.4 x 10-19 = (19.89 x 10-26)/λ

∴ 10.61 x 10-19  = (19.89 x 10-26)/λ

∴ λ  = (19.89 x 10-26)/(10.61 x 10-19)



∴ λ  = 1.871 x 10-7  = 1871x 10-10 m = 1871 Å

Ans: The wavelength of radiation is 1871 Å

Example – 01:

  • Radiation of wavelength 3000 A falls on a photoelectric surface for which work function is 1.6 eV. What is the stopping potential for emitted electron.
  • Given:  Wavelength of radiation = λ = 3000 Å = 3000 x 10-10 m, work function = Φ = 1.6 eV = 1.6 x 1.6 x 10-19 J = 2.56 x 10-19 J, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
  • To Find: Stopping potential = Vs =?
  • Solution:

By Einstein’s photoelectric equation

K.E. max = hν – Φ  = hc/λ – Φ

But K.E. max = eVs



 eVs =  hc/λ – Φ

∴ (1.6 x 10-19) x Vs = (6.63 x 10-34)(3 x 108)/(3000 x 10-10) – 2.56 x 10-19

∴ (1.6 x 10-19) x Vs = 6.63 x 10-19 – 2.56 x 10-19

∴  Vs = 4.07 x 10-19

∴ Vs  = (4.07 x 10-19)/(1.6 x 10-19)



∴ Vs  = 2.54 V

Ans: The stopping potential is 2.54 V.

Example – 03:

  • When a radiation of certain wavelength shines on the cathode of the photoelectric cell, the photocurrent produced can be reduced to zero by applying stopping potential of 3 V. If the work function of the photoemitter is 3.63 eV, find the frequency of radiation.
  • Given: Stopping potential = Vs = 3 V, work function = Φ = 3.63 eV = 3.63 x 1.6 x 10-19 J, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
  • To Find: frequency of radiation = ν = ?
  • Solution:

By Einstein’s photoelectric equation

K.E. max = hν – Φ  = hc/λ – Φ

But K.E. max = eVs



 eVs =  hν – Φ

∴ (1.6 x 10-19) x(3) = (6.63 x 10-34) ν – 3.63 x 1.6 x 10-19

∴ 4.8 x 10-19 = (6.63 x 10-34) ν – 5.808 x 10-19

∴ 4.8 x 10-19 +  5.808 x 10-19 = (6.63 x 10-34) ν



∴ 10.608 x 10-19 = (6.63 x 10-34) ν

∴ ν  = (10.61 x 10-19)/(6.63 x 10-34)

∴ ν  = 1.6 x 1015  Hz

Ans: The frequency of radiation is 1.6 x 1015  Hz

Example – 04:

  • Photoelectrons emitted by a surface have maximum kinetic energy of 4 x 10-19 J. What is the stopping potential for phpto emission from the surface for the incident radiation.
  • Given: Maximum kinetic energy of photoelectron = K.E. max = 4 x 10-19 J, Charge on electron = e = 1.6 x 10-19 C.
  • To Find: frequency of radiation = ν = ?
  • Solution:

K.E. max = eVs

Vs =  K.E. max/e



∴   Vs =  (4 x 10-19)/(1.6 x 10-19) = 2.5 V

Ans: The stopping potential = 2.5 V

Example – 05:

  • Light of wavelength 2000 Å is incident on the cathode of a photocell. The current in the photocell is reduced to zero by stopping potential of 2 V. Find the threshold wavelength of the material of cathode.
  • Given: Stopping potential = Vs = 2 V, wavelength of incident light = λ = 2000 Å = 2000 x 10-10 m, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
  • To Find: frequency of radiation = ν = ?
  • Solution:

By Einstein’s photoelectric equation

K.E. max = hν – Φ  = hc/λ – Φ



But K.E. max = eVs

eVs =  hc/λ – Φ

∴ (1.6 x 10-19) x 2 = (6.63 x 10-34)(3 x 108)/(2000 x 10-10)- Φ

∴3.2 x 10-19 = 9.945 x 10-19 –  Φ

∴ Φ =  9.945 x 10-19 –  3.2 x 10-19 = 6.745 x 10-19 J

We have Φ = h νo  = hc/λo



∴  λo = hc/Φ =(6.63 x 10-34) x (3 x 108) / ( 6.745 x 10-19) = 2.949 x 10-7 m

∴  λo= 2949 x 10-10 m = 2949 Å

Ans: The threshold  wavelength is 2949 Å

Example – 06:

  • Photoelectrons are ejected fro metal surface when radiation of wavelength 160 nm is incident on the surface. Find stopping potential of emitted electrons if the limiting wavelength is 240 nm for photoelectric emission from the surface.
  • Given: Stopping potential = Vs = 2 V, wavelength of incident light = λ = 160 nm = 160 x 10-9 m, Threshold wavelength = λo = 240 nm = 240 x 10-9 m, , speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
  • To Find:Stopping potential = Vs =?
  • Solution:

By Einstein’s photoelectric equation

K.E. max = hν – Φ  = hc/λ – hc/λo

But K.E. max = eVs

∴ eVs = hc/λ – hc/λo

∴ eVs = hc(1/λ – 1/λo)

∴ (1.6 x 10-19) x Vs = (6.63 x 10-34)(3 x 108)(1/(160 x 10-9) – 1/(240 x 10-9))

∴ (1.6 x 10-19) x Vs = (6.63 x 10-34)(3 x 108)x 10x (1/160  – 1/240 )

∴ (1.6 x 10-19) x Vs = 19.89 x 10-17 x (240 -160)/(160 x240 )

∴ (1.6 x 10-19) x Vs = 19.89 x 10-17 x 80/(160 x240 )

∴ (1.6 x 10-19) x Vs = 4.143 x 10-19

∴  Vs = 4.143 x 10-19/  (1.6 x 10-19) = 2.59 eV

Ans: the stopping potential is 2.59 eV

Example – 07:

  • Calculate the change in stopping potential when the wavelength of light incident on photoelectric surface is reduced from 4000 Å to 3600 Å.
  • Given: Initial wavelength = λ1 = 4000 Å  = 4000 x 10-10 m = 4 x 10-7 m, Final wavelength = λ2 = 3600 Å = 3600 x 10-10 m = 3.6 x 10-7  m, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
  • To Find:Stopping potential = Vs =?
  • Solution:

By Einstein’s photoelectric equation

K.E. max = hν – Φ  = hc/λ – Φ

But K.E. max = eVs

∴ eVs1 = hc/λ1 – Φ  ……….. (1)

∴ eVs2 = hc/λ2 – Φ  ……….. (2)

Subtracting equation (1) from (2)

∴ eVs2 – eVs1 = (hc/λ2 – Φ) – (hc/λ1 – Φ)

∴ e(Vs2 – Vs1) = hc/λ2 – hc/λ1

∴ e(Vs2 – Vs1) = hc(1/λ2 – 1/λ12)

∴ (1.6 x 10-19) (Vs2 – Vs1) = (6.63 x 10-34)(3 x 108)(1/(3.6x 10-7) – 1/(4 x 10-7))

∴ (1.6 x 10-19) (Vs2 – Vs1) = (6.63 x 10-34)(3 x 108) x 10x (1/3.6 – 1/4)

∴ (1.6 x 10-19) (Vs2 – Vs1) = 19.89 x 10-19 x (4 – 3.6)/1(3.6 x 1/4)

∴ (1.6 x 10-19) (Vs2 – Vs1) = 19.89 x 10-19 x 0.4/1(3.6 x 1/4)

∴ (1.6 x 10-19) (Vs2 – Vs1) = 5.525 x 10-20 

∴  (Vs2 – Vs1) = 5.525 x 10-20/(1.6 x 10-19) = 0.3453 V

Ans: the change in stopping potential is 0.3453 eV

Example – 08:

  • When light of frequency 2.2 x 1015 Hz is incident on a metal surface, photoelectric current can be stopped by a retarding potential of 6.6 V. For light of frequency 4.6 x 1015 Hz the reverse potential is 16.5 V. Find h
  • Given: Initial frequency = ν1 = 2.2 x 1015 Hz, initial stopping potential = Vs1 =6.6 V, Final frequency = ν2 = 4.6 x 1015 Hz, Final stopping potential = Vs2 = 16.5 V, Charge on electron = e = 1.6 x 10-19 C.
  • To Find: Planck’s constant = h = ?
  • Solution:

By Einstein’s photoelectric equation

K.E. max = hν – Φ

But K.E. max = eVs

∴ eVs = hν – Φ

∴ eVs1 = hc/λ1 – Φ  ……….. (1)

∴ eVs2 = hc/λ2 – Φ  ……….. (2)

Subtracting equation (1) from (2)

∴ eVs2 – eVs1 = (hν2 – Φ) – (hν1 – Φ)

∴ e(Vs2 – Vs1) = hν2 – hν1

∴ e(Vs2 – Vs1) = h(ν2 – ν2)

∴ (1.6 x 10-19) (16.5 – 6.6) = h(4.6 x 1015 – 2.2 x 1015)

∴ 1.6 x 10-19  x 9.9 = h(2.4 x 1015)

∴  h = (1.6 x 10-19  x 9.9)/(2.4 x 1015)

∴  h = 6.6 x 10-34  Js

Ans: the value of Planck’s constant is 6.6 x 10-34  Js

Example – 09:

  • When light of frequency 2 x 1015 Hz is incident on a metal surface, photoelectric current can be stopped by a retarding potential of 6 V. For light of frequency 1015 Hz the reverse potential is 2 V. Find Planck’s constant, work function and threshold frequency.
  • Given: Initial frequency = ν1 = 2 x 1015 Hz, initial stopping potential = Vs1 = 6 V, Final frequency = ν2 = 1015 Hz, Final stopping potential = Vs2 = 2 V, speed of light = c = 3 x 108 m/s, Charge on electron = e = 1.6 x 10-19 C.
  • To Find: Planck’s constant = h = ?
  • Solution:

By Einstein’s photoelectric equation

K.E. max = hν – Φ

But K.E. max = eVs

∴ eVs = hν – Φ

∴ eVs1 = hν1 – Φ  ……….. (1)

∴ eVs2 = hν2 – Φ  ……….. (2)

Subtracting equation (2) from (1)

∴ eVs1 – eVs2 = (hν1 – Φ) – (hν2 – Φ)

∴ e(Vs1 – Vs2) = hν1 – hν2

∴ e(Vs1 – Vs2) = h(ν1 – ν2)

∴ (1.6 x 10-19) (6 – 2) = h(2 x 1015 – 1015)

∴ 1.6 x 10-19  x 4 = h(1 x 1015)

∴  h = (1.6 x 10-19  x 4)/1 x 1015)

∴  h = 6.4 x 10-34  Js

From equation (1) we have

eVs1 = hν1 – Φ

∴  1.6 x 10-19 x 6 = 6.4 x 10-34 x 2 x 1015 – Φ

∴  9.6 x 10-19  = 12.8 x 10-19 – Φ

∴  Φ = 12.8 x 10-19 – 9.6 x 10-19

∴  Φ = 3.2 x 10-19  J

∴  Φ = (3.2 x 10-19)/(1.6 x 10-19) = 2 eV

Φ = hνo

νo = Φ /h = (3.2 x 10-19 )/(6.4 x 10-34) = 5 x 1014 Hz

Ans: the value of Planck’s constant is 6.4 x 10-34  Js, work function = 2 eV

and threshold frequency = 5 x 1014 Hz

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