# Problems on Energy of Photon and Work Function

 Science > 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 > You are Here

1. Martin Kenamil
• Hemant More