Easiest explanation of Davisson and Germer experiment class 12

Easiest explanation of Davisson and Germer experiment class 12


  1. In today’s article, we are going to study about ‘  Easiest explanation of Davisson and Germer experiment class 12′ How beneficial it is and what is the Experimental setup of Davisson and germer experiment. We will also read about some more topics like explanation of Davisson and Germer experiment, conclusion of Davisson and Germer experiment, numerical based on Davisson and Germer experiment etc in Detail.  So, take your notebooks in you hand and get ready to study physics in an easy and sorted way.

Davisson and Germer experiment

Wave nature of particle was verified by davisson and germer experiment. The experimental figure is shown below. This experimental setup has three parts-

  1. Electron gun- It consist a tungsten filaments coated by barium oxide. The electron gun emits electron by giving power to the tungsten filament using law voltage power supply. A drift tube ‘A’ is used for focusing and accelerating electron beam. The energy of electron is controlled by high voltage power supply. This power supply apply a voltage difference between filament and tube A.
  2. Nickel Target- An nickel crystal is used as a target for electron beam.
  3. Movable collector- It is a hollow chamber filled CO2 or SO2 gas. This collector is connected to a galvanometer for measurement of current.

Working of Davisson and Germer experiment

A electron beam is generated by a electron gun. The energy of electron beam is controlled by applying a fix voltage v. When this electron beam is diffracted by nickel crystal. Intensity of diffracting beam is measured by a movable collecter that is connected connected to a galvanometer. This experiment was repeated by varying potential value from 44v to 69v. And measure the intensity of beam at different angle.

If we plot a graph between direction of electron beam and potential v then we get at 54v. We get maximum pitch at 50° angle.


As we know from De-broglie principle,

λ= h/ √2mev

On putting the value of e,h,m and v then we get

λ= 12.27× 10^-10 / √v

λ= 12.27× 10^-10 m/ √54

And hence, λ= 1.65× 10^-10 m

λe = 1.65× 10^-10 × 10^9 nm

λe = 1.65× 10^-1

And λe = 0.165 nm

We know that for diffraction of wave-

nλ = d sinθ

If n= 1 and 2.15 A° because we takes known crystal (Nickel)

λ = d sinθ

λ = 2.15 × 10^-10 × sin 50°

And λ= 1.67× 10^-10 m

∴ sin θ= 0.766

λe = 1.67A°

λe = 0.167 nm

And, λe = 1.67× 10^-10 m

Above all these three values are Experimental value.

Numerical based on Davisson and Germer experiment

Question 1 If intermolecular distance (d) is 4A° for a crystal if Maximum diffraction of beam is observed at 30° then find out its wavelength.

Solution- given that θ= 30° and d= 4A°

λ= ?

We know that,

λ = d sinθ

Hence on putting values we get,

λ = 4× 10^-10 × sin30°

λ = 4× 10^-10 × 1/2

And , λ = 2× 10^-10 m

Question 2- If intermolecular distance is 3A° for a crystal. A electron bean falling upon this crystal. Electron beam is given 1.5A° then find out interference angle.

Solution- Given that d= 3A °

λ = 1.5 A°

To find θ = ?

Hence, we know that

λ = d sinθ

Sin θ = λ/d

Sin θ = 1.5 A°/ 3A°

And, Sin θ = 15/30 = 1/2

Hence θ is 30°.


Leave a Reply

Your email address will not be published.