Numerical Problems on Photoelectric Effect

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In this article, we shall study to calculate, Energy of incident photon, threshold wavelength and threshold frequency of metal.

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 wavelength of photon is 41.44 Å

Example – 04:

Find the energy of a photon in eV if its wavelength is 10 m

Given: Wavelength of photon = 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:

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 = ν_o =? Threshold wavelength of silver = λ_o = ?

Solution:

We have Φ = h ν_o

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

Now c = ν_o λ_o

∴ λ_o = c/ν_o = (3 x 10⁸)/( 8 x 10¹⁴) = 3.750 x 10^-7 m

∴ λ_o= 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 = λ_o = 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 ν_o = hc/λ_o

∴    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 the lowest 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 = ν_o =?

Solution:

We have Φ = h ν_o

∴   ν_o = 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 – 09:

The photoelectric work function of platinum is 6.3 eV and the 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 ν_o = hc/λ_o

∴   h = Φλ_o/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 Planck’s constant is 6.625 x 10^-34 Js

Example – 10:

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 = λ_o =?

Solution:

We have Φ = h ν_o = hc/λ_o

∴ λ_o = 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

∴ λ_o= 9418 x 10^-10 m = 9418 Å

Ans: The threshold wavelength is 9418 Å

Example – 11:

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 = λ_o =?

Solution:

We have Φ = h ν_o

∴   ν_o = 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 – 12:

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 = λ_o =?

Solution:

We have Φ = h ν_o

∴   ν_o = 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 – 13:

The photoelectric work function of a metal is 3 eV. Calculate the threshold frequency for the metal. If 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 = λ_o =?

Solution:

We have Φ = h ν_o

∴   ν_o = 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.

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Science > Physics > Photoelectric Effect > Numerical Problems on Photoelectric Effect