Problems on Energy of Photon and Work Function

Physics Chemistry Biology Mathematics
Science > Physics > Electrons and Photons > You are Here

Example – 01:

  • The energy of a photon is 2.59 eV. Find its frequency and wavelength.
  • Given: Energy of photon = E = 2.59 eV = 2.59 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
  • To Find: Frequency of photon = ν = ?, Wavelength = λ = ?
  • Solution:

We have E = h ν

∴   ν = E/h = (2.59 x 1.6 x 10-19)/(6.63 x 10-34) = 6.244 x 1014 Hz

Now c = ν λ

∴  λ = c/ν = (3 x 108)/( 6.244 x 1014) = 4.805 x 10-7 m



∴  λ = 4805 x 10-10 m = 4805 Å

Ans: The frequency of photon is 6.244 x 1014 Hz and its wavelength is 4805 Å

Example – 02:

  • The energy of a photon is 1.0 x 10-8 J. Find its frequency and wavelength.
  • Given: Energy of photon = E = 1.0 x 10-18 J, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js
  • To Find: Frequency of photon = ν = ?, Wavelength = λ = ?
  • Solution:

We have E = h ν

∴   ν = E/h = (1.0 x 10-18)/(6.63 x 10-34) = 1.508 x 1015 Hz



Now c = ν λ

∴  λ = c/ν = (3 x 108)/( 1.508 x 1015) = 1.989 x 10-7 m

∴  λ = 1989 x 10-10 m = 1989 Å

Ans: The frequency of photon is 1.508 x 1014 Hz and its wavelength is 1989 Å

Example – 03:

  • The energy of a photon is 300 eV. Find its wavelength.
  • Given: Energy of photon = E = 300 eV = 300 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
  • To Find: Wavelength = λ = ?
  • Solution:

We have E = h ν = hc/λ



∴  λ = hc/E = (6.63 x 10-34)(3 x 108)/(300 x 1.6 x 10-19) = 4.144 x 10-9 m

∴  λ = 41.44 x 10-10 m = 41.44 Å

Ans: The wavelengthof photon is 41.44 Å

Example – 04:

  • Find the energy of a photon in eV if its wavelength is 10 m
  • Given: Wavelength of photopn = λ = 10 m, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js
  • To Find: Wavelength = λ = ?
  • Solution:

We have E = h ν = hc/λ

∴  E  = hc/λ = (6.63 x 10-34)(3 x 108)/(10) = 19.89 x 10-27 J



∴  E  = (19.89 x 10-27)/(1.6 x 10-19) = 1.243 x 10-7 eV

Ans: The energy of the photon is 1.243 x 10-7 eV

Example – 05:

  • Find the energy of a photon whose frequency is 5.0 x 1014 Hz
  • Given: Frequency of photon = ν = 5.0 x 1014 Hz, Planck’s constant = h = 6.63 x 10-34 Js
  • To Find: Energy of photon = E = ?
  • Solution:

We have E = h ν

∴  E = (6.63 x 10-34) x (5.0 x 1014)= 3.315 x 10-29  J

Ans: The energy of the photon is 3.315 x 10-29  J



Example – 06:

  • The photoelectric work function of silver is 3.315 eV. calculate the threshold frequency and threshold wavelength of silver.
  • Given: Work function of silver = Φ = 3.315 eV = 3.315 x 1.6 x 10-19 J, speed of light = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js
  • To Find: Threshold frequency of silver = νo = ?, Threshold wavelength of silver = λo = ?
  • Solution:

We have Φ = h νo

∴   νo = Φ/h = (3.315 x 1.6 x 10-19)/(6.63 x 10-34) = 8 x 1014 Hz

Now c = νo λo

∴  λo = c/νo = (3 x 108)/( 8 x 1014) = 3.750 x 10-7 m



∴  λo= 3750 x 10-10 m = 4805 Å

Ans: The threshold  frequency of silver is 8 x 1014 Hz and its threshold  wavelength is 3750 Å

Example – 07:

  • A light of wavelength 4800 Å can just cause photoemission from a metal. What is the photoelectric work function for metal in eV?
  • Given: Threshold wavelength = λo = 4800 Å = 4800 x 10-10 m, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js
  • To Find: Work function of silver = Φ  = ?
  • Solution:

We have Φ = h νo  = hc/λo

∴    Φ = (6.63 x 10-34)x(3 x 108)/(4800 x 10-10) = 4.144 x 10-19 J

∴    Φ =  (4.144 x 10-19)/(1.6 x 10-19) = 2.59 eV

Ans: The photoelectric work function of the metal is 2.59 eV



Example – 08:

  • The photoelectric work function of a metal is 2 eV. calculate thelowest frequency radiation that will cause photoemission from the surface.
  • Given: Work function of silver = Φ = 2 eV = 2 x 1.6 x 10-19 J, Planck’s constant = h = 6.63 x 10-34 Js
  • To Find: Threshold frequency of silver = νo = ?,
  • Solution:

We have Φ = h νo

∴   νo = Φ/h = (2 x 1.6 x 10-19)/(6.63 x 10-34) = 4.827 x 1014 Hz

Ans: The threshold  frequency of metal is 4.827 x 1014.

Example – 08:

  • The photoelectric work function of platinum is 6.3 eV and longest wavelength that can eject photoelectron from platinum is 1972 Å. Calculate the Planck’s constant.
  • Given: Work function of platinum = Φ = 6.3 eV =6.3 x 1.6 x 10-19 J, Threshold wavelength of silver = 1972 Å = 1972  x 10-10 m, speed of light = c = 3 x 108 m/s,
  • To Find: Planck’s constant = h = ?,
  • Solution:

We have Φ = h νo  = hc/λo



∴   h = Φλo/c = (6.3 x 1.6 x 10-19) x (1972 x 10-10)/(3 x 108) = 6.625 x 10-34 Js

Ans: The value of Palnck’s constant is  6.625 x 10-34 Js

Example – 09:

  • The photoelectric work function of metal is 1.32 eV. calculate the longest wavelength that can cause photoelectric emission from the metal surface.
  • Given: Work function of silver = Φ = 1.32 eV = 1.32 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
  • To Find: Threshold wavelength of metal = λo = ?
  • Solution:

We have Φ = h νo  = hc/λo

∴  λo = hc/Φ =(6.63 x 10-34) x (3 x 108) / (1.32 x 1.6 x 10-19) = 9.418 x 10-7 m

∴  λo= 9418 x 10-10 m = 9418 Å

Ans: The threshold  wavelength is 9418 Å



Example – 10:

  • The photoelectric work function of metal is 5 eV. calculate the threshold frequency for the metal. If a light of wavelength 4000 Å is incident on this metal surface, will photoelectron will be ejected?
  • Given: Work function of silver = Φ = 5 eV = 5 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, wavelength of incident light = λ = 4000 Å = 4000  x 10-10 m
  • To Find: Threshold wavelength of metal = λo = ?
  • Solution:

We have Φ = h νo

∴   νo = Φ/h = (5 x 1.6 x 10-19)/(6.63 x 10-34) = 1.2 x 1015 Hz

Now c = ν λ

∴  ν = c/λ = (3 x 108)/( 4000 x 10-10) = 7.5 x 1014 Hz

The frequency of incident light is less than the threshold frequency.

No photoelectrons will be emitted from the metal surface.

Ans: The threshold frequency is 1.2 x 1015 Hz and no photoelectron will be emitted.

Example – 11:

  • The photoelectric work function of a metal is 2.4 eV. calculate the incident frequency, the threshold frequency for the metal. If a light of wavelength 6800 Å is incident on this metal surface, will photoelectron will be ejected?
  • Given: Work function of silver = Φ = 2.4 eV= 2.4 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, wavelength of incident light = λ = 6800 Å = 6800  x 10-10 m
  • To Find: Threshold wavelength of metal = λo = ?
  • Solution:

We have Φ = h νo

∴   νo = Φ/h = (2.4 x 1.6 x 10-19)/(6.63 x 10-34) = 5.79 x 1014 Hz

Now c = ν λ

∴  ν = c/λ = (3 x 108)/( 6800 x 10-10) = 4.41 x 1014 Hz

The frequency of incident light is less than the threshold frequency.

No photoelectrons will be emitted from the metal surface.

Ans: The incident frequency is 4.41 x 1014 Hz and the threshold frequency is 5.79 x 1014 Hz,

and no photoelectron will be ejected.

Example – 12:

  • The photoelectric work function of a metal is 3 eV. calculate the threshold frequency for the metal. If a light of wavelength 6000 Å is incident on this metal surface, will photoelectron will be ejected?
  • Given: Work function of silver = Φ = 3 eV = 3 x 1.6 x 10-19 J, speed of light = 3 x 108 m/s, planck’s constant = h = 6.63 x 10-34 Js, wavelength of incident light = λ = 6000 Å = 6000  x 10-10 m
  • To Find: Threshold wavelength of metal = λo = ?
  • Solution:

We have Φ = h νo

∴   νo = Φ/h = (3 x 1.6 x 10-19)/(6.63 x 10-34) = 7.24 x 1014 Hz

Now c = ν λ

∴  ν = c/λ = (3 x 108)/( 6000 x 10-10) = 5 x 1014 Hz

The frequency of incident light is less than the threshold frequency.

No photoelectrons will be emitted from the metal surface.

Ans: The threshold frequency is 7.24 x 1014 Hz,

and no photoelectron will be ejected.

Science > Physics > Electrons and Photons > You are Here
Physics Chemistry Biology Mathematics

2 Comments

  1. Martin Kenamil

    Thank you

    • Thank you Martin. Our aim is to deliver knowledge in the simplest language so that everybody can understand it. If you like the website spread it through social media in your groups. That is the way we can reach maximum people.

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