Q.(a) Describe Davisson-Germer experiment which demonstrated the wave nature of electrons. OR

(a) The Davisson-Germer experiment found a sharp scattered-electron intensity peak matching de Broglie's predicted diffraction wavelength, confirming electron wave nature; (b) Bohr's energy formula gives $E_n=-13.6/n^2$ eV, and for $E=-3.4$ eV ($n=2$), the angular momentum is $L=h/\pi\approx2.11\times10^{-34}$ J s. Both alternatives answered below.

(a) Davisson-Germer experiment

Apparatus. An electron gun (heated filament + accelerating potential $V$) produces a fine beam of electrons directed at a nickel crystal target. A movable detector, positioned at various scattering angles $\theta$ from the incident beam, measures the intensity of electrons scattered by the crystal.

Observation. For ordinary (non-crystalline) scattering, intensity would vary smoothly with angle. Instead, at an accelerating voltage of $54$ V, a pronounced, sharp peak in scattered electron intensity was observed at a scattering angle of $50°$ — behaviour characteristic of diffraction, not simple particle scattering.

Interpretation. The nickel crystal's regularly spaced atomic planes act like a diffraction grating. Using Bragg's law for the observed diffraction peak, the wavelength associated with the 54 V electrons works out to about $1.65$ Å.

Comparison with de Broglie's hypothesis. For an electron accelerated through 54 V, the predicted de Broglie wavelength is

$\lambda=\dfrac{h}{\sqrt{2meV}}\approx1.67\ \text{Å}$

which matches the experimentally observed value ($1.65$ Å) very closely.

This agreement provided direct, quantitative experimental confirmation that electrons — usually thought of as particles — exhibit wave-like diffraction behaviour, verifying de Broglie's hypothesis.

(b)(i) Orbital energy of the electron in hydrogen atom (Bohr theory)

For an electron of charge $-e$ orbiting a nucleus of charge $+e$ in the $n^{th}$ Bohr orbit of radius $r_n$:

Coulomb attraction provides the centripetal force:

$\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r_n^2}=\dfrac{mv_n^2}{r_n}\ \Rightarrow\ KE=\dfrac12mv_n^2=\dfrac{e^2}{8\pi\varepsilon_0r_n}$

Potential energy:

$PE=-\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r_n}$

Total energy:

$E_n=KE+PE=\dfrac{e^2}{8\pi\varepsilon_0r_n}-\dfrac{e^2}{4\pi\varepsilon_0r_n}=-\dfrac{e^2}{8\pi\varepsilon_0r_n}$

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