Question

Davisson and Germer performed their experiment with a nickel target for several electron bombarding energies. At what angles would they find diffraction maxima for 38 eV and 54 eV electrons?

Answer 1

Answer:

The angle of diffraction are 67.75 deg and 53.57 deg.

Explanation:

Given:

Davisson and Germer  experiment with nickel target for electrons bombarding.

Voltages : $38\ eV$ and $54\ eV$

We have to find the angles that is  $\phi_3_8$ and $\phi_5_4$ .

Concept:

• Davison Germer experiment is based on de Broglie hypothesis where it says matter has both wave and particle nature.
• When electrons get reflected from the surface of a metal target with an atomic spacing of $D$, they form diffraction patterns.
• The positions of diffraction maxima are given by $Dsin(\phi) = n\lambda$ .
• An atomic spacing is $D = 0.215\ nm$, when  the principal maximum corresponds to n=1
• The wavelength is $\lambda$, and   $\lambda =\frac{1.227 \sqrt{V} } {\sqrt{V_o}}\ nm$ .

Solution:

Finding the wavelength at $V_o=38\ eV$ .

⇒ $\lambda_3_8 =\frac{1.227 \sqrt{V} } {\sqrt{38V}}\ nm$

⇒ $\lambda_3_8 =0.199$ nm

    Plugging the values of wavelength.

⇒  $sin(\phi)=\frac{\lambda}{D}$

⇒  $\phi_3_8=sin^-1(\frac{0.199}{0.215} )$

⇒ $\phi_3_8 =67.75$ degrees.

Now

For for the electrons with energy $54\ eV$, $V_0=54V$ the wavelength is.

⇒ $\lambda_5_4 =\frac{1.227 \sqrt{V} } {\sqrt{54V}} = 0.173$ nm

And

⇒ $\phi_5_4=sin^-1(\frac{0.173}{0.215} ) = 53.57$ degrees.

So,

The angles of diffraction maxima are 67.75 deg and 53.57 deg.