JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Davision and Germer Experiment

(1) It is used to study the scattering of electron from a solid or to verify the wave nature of electron. A beam of electrons emitted by electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. Ni crystal behaves like a three dimensional diffraction grating and it diffracts the electron beam obtained from electron gun.

(2) The diffracted beam of electrons is received by the detector which can be positioned at any angle by rotating it about the point of incidence. The energy of the incident beam of electrons can also be varied by changing the applied voltage to the electron gun.

(3) According to classical physics, the intensity of scattered beam of electrons at all scattering angle will be same but Davisson and Germer, found that the intensity of scattered beam of electrons was not the same but different at different angles of scattering. It is maximum for diffracting angle 50° at 54 volt potential difference.

(4) If the de-Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg's formula \[2d\sin \theta =n\lambda \], we can determine the wavelength of these waves.

where d = distance between diffracting planes, \[\theta =\frac{(180-\varphi )}{2}\]= glancing angle for incident beam = Bragg's angle.

The distance between diffracting planes in Ni-crystal for this experiment is \[d=0.91\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\] and the Bragg's angle \[={{65}^{o}}\]. This gives for \[n=1,\,\,\lambda =2\times 0.91\times {{10}^{-10}}\,\sin {{65}^{o}}=1.65\overset{\text{o}}{\mathop{\text{A}}}\,\]

Now the de-Broglie wavelength can also be determined by using the formula \[\lambda =\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{54}}=1.67{\AA}\]. Thus the de-Broglie hypothesis is verified.

(5) The Bragg's formula can be rewritten in the form containing interatomic distance D and angle \[\phi \]

\[\because \]\[\theta =90-\frac{\varphi }{2}\,\,\,\text{and }\,d=D\cos \theta =D\sin \frac{\varphi }{2}\]

Using \[\sin \theta =\cos \frac{\varphi }{2}\] \[2d\sin \theta =\lambda \]\[\Rightarrow

\]\[2(D\sin \frac{\varphi }{2}).\cos \frac{\varphi }{2}=\lambda \]\[\Rightarrow \]\[D\,\,\,\sin \,\,\phi =\lambda \]