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Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list:
List-I List-II
A. Franck-Hertz Experiment (i) Particle nature of light
B. Photo-electric experiment (ii) Discrete energy levels of the atom
C. Davison-Germer Experiment (iii) Wave nature of electron
(iv) Structure of atom
We begin by recalling what each classic experiment in List-I actually established in the language of modern physics.
First, the Franck-Hertz experiment bombarded mercury atoms with electrons of variable kinetic energy. The observed current showed sharp drops only at certain fixed electron energies. Those fixed values matched the energy difference between the ground state and higher states of the mercury atom. This proved that an atom can absorb energy only in fixed packets. In other words, it revealed the discrete energy levels of the atom
$$A \;\; \longrightarrow \;\; (ii).$$
Next, the photo-electric experiment illuminated a metal surface with light of various frequencies and measured the emitted electrons. The key observations were: (a) there is a threshold frequency $$\nu_0$$ below which no electrons are ejected, and (b) above this threshold, the maximum kinetic energy $$K_{\max}$$ of the emitted electrons grows linearly with the light frequency $$\nu$$, obeying
$$K_{\max}=h\nu-\phi,$$
where $$h$$ is Planck’s constant and $$\phi$$ is the work function of the metal. This linear relation can be explained only if light arrives in discrete packets of energy $$E=h\nu$$, i.e. photons, demonstrating the particle nature of light
$$B \;\; \longrightarrow \;\; (i).$$
Finally, the Davisson-Germer experiment directed a beam of electrons at a nickel crystal and recorded the intensity of electrons scattered at various angles. A strong intensity peak appeared only at particular angles that satisfied Bragg’s law for diffraction, namely
$$2d\sin\theta = n\lambda,$$
where $$d$$ is the inter-planar spacing of the crystal, $$\theta$$ is the glancing angle, $$n$$ is an integer, and $$\lambda$$ is the wavelength of the incident waves. Inserting the de Broglie relation $$\lambda = \dfrac{h}{p}$$ with $$p$$ the electron momentum reproduced exactly the observed angles. Those diffraction fringes are possible only if electrons possess wave character. Hence this experiment confirmed the wave nature of the electron
$$C \;\; \longrightarrow \;\; (iii).$$
Collecting the three correspondences we have
$$A\!-\!(ii), \qquad B\!-\!(i), \qquad C\!-\!(iii).$$
Examining the options, Option D lists precisely this mapping. Hence, the correct answer is Option D.
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