Davisson-Germer Experiment and Heisenberg Uncertainty Principle Questions in English

(D) The Davisson and Germer experiment provided the first experimental evidence for the wave nature of matter,specifically electrons. By observing the diffraction of electrons from a nickel crystal,they confirmed the de Broglie hypothesis,which states that particles like electrons exhibit wave-like properties. Since none of the options $(a)$,$(b)$,or $(c)$ correctly describe this,and the provided option $(d)$ was corrected to reflect the actual scientific outcome,the correct answer is $(d)$.

(C) The Davisson-Germer experiment was the first to provide experimental evidence for the wave nature of matter,as proposed by de Broglie.
In this experiment,a beam of electrons is accelerated through a potential difference and directed at a nickel $(Ni)$ crystal.
The electrons scatter off the crystal lattice,and the intensity of the scattered electrons is measured at different angles.
The observed diffraction pattern confirms that electrons behave as waves,exhibiting interference and diffraction phenomena similar to light or $X$-rays.
Therefore,the waves that are diffracted in this experiment are electron waves.

(A) According to the Bragg's law for diffraction,the condition for the first maximum is given by $d \sin \theta = n \lambda$.
For the first maximum,$n = 1$,so $d \sin \theta = \lambda$.
Given: Wavelength $\lambda = 1.5 \ \mathring{A}$ and atomic spacing $d = 3 \ \mathring{A}$.
Substituting the values: $3 \sin \theta = 1.5$.
Therefore,$\sin \theta = \frac{1.5}{3} = 0.5$.
Thus,$\theta = \arcsin(0.5) = 30^{\circ}$.

(B) In the Davisson-Germer experiment,the electron beam is directed at a nickel crystal. The atoms in the crystal lattice act as scattering centers,and the crystal structure behaves as a three-dimensional diffraction grating for the incident electron waves,confirming the wave nature of matter.

(B) The de Broglie wavelength $\lambda$ of an electron with energy $E = 60 \ eV$ is given by $\lambda = \frac{12.27}{\sqrt{E}} \mathring A = \frac{12.27}{\sqrt{60}} \approx 1.58 \mathring A$.
In the Davisson-Germer experiment,the scattering angle $\phi$ is the angle between the incident and scattered beam. The glancing angle $\theta$ is related to $\phi$ by $\theta = 90^\circ - \frac{\phi}{2}$.
Given $\phi = 50^\circ$,we have $\theta = 90^\circ - 25^\circ = 65^\circ$.
Using Bragg's Law for diffraction,$2d \sin \theta = n\lambda$. For the first order $(n=1)$,$d = \frac{\lambda}{2 \sin \theta}$.
Substituting the values: $d = \frac{1.58}{2 \sin 65^\circ} = \frac{1.58}{2 \times 0.906} \approx 0.87 \mathring A$.
Note: Based on standard textbook values for this specific experiment where $\phi = 50^\circ$ and $E = 54 \ eV$,the spacing $d$ is approximately $0.91 \mathring A$. Given the options provided,$0.9 \mathring A$ is the correct choice.

(B) The Bragg's law for diffraction is given by $2d \sin \theta = n\lambda$,where $\theta$ is the glancing angle.
From the geometry of the problem,the angle of incidence $i$ is measured with respect to the normal to the crystal plane. Thus,the glancing angle $\theta = 90^\circ - i = 90^\circ - 30^\circ = 60^\circ$.
Assuming first-order diffraction $(n = 1)$,we have $2d \sin(60^\circ) = \lambda$.
Substituting $d = 1 \ \mathring{A}$,we get $\lambda = 2 \times 1 \times \frac{\sqrt{3}}{2} = \sqrt{3} \ \mathring{A} \approx 1.732 \ \mathring{A}$.
The de Broglie wavelength of an electron accelerated by potential $V$ is given by $\lambda = \sqrt{\frac{150}{V}} \ \mathring{A}$.
Equating the two: $\sqrt{3} = \sqrt{\frac{150}{V}}$.
Squaring both sides: $3 = \frac{150}{V}$.
Therefore,$V = \frac{150}{3} = 50 \ V$.

(B) The Bragg's law for diffraction is given by $2d\sin\theta = n\lambda$,where $\theta$ is the glancing angle (the angle between the incident ray and the crystal plane).
From the given figure,the angle of incidence $i$ is measured with respect to the normal to the crystal plane.
Therefore,the glancing angle $\theta$ is related to the angle of incidence $i$ by $\theta = 90^\circ - i$.
Substituting this into Bragg's law:
$2d\sin(90^\circ - i) = n\lambda$
Since $\sin(90^\circ - i) = \cos i$,the equation becomes:
$2d\cos i = n\lambda$.

(D) The Davisson-Germer experiment demonstrates the wave nature of electrons through electron diffraction,not interference.
$1$. The nickel crystal acts as a three-dimensional diffraction grating for electrons.
$2$. The interatomic spacing in the nickel crystal is comparable to the de Broglie wavelength of the electrons used.
$3$. The electron gun produces a beam of electrons with a specific,constant energy.
$4$. Since the experiment is based on diffraction,option $D$ is the incorrect statement.

(A) The Davisson-Germer experiment demonstrated the wave nature of electrons by observing electron diffraction.
In this experiment,a beam of electrons was directed at a nickel crystal.
The intensity of the scattered electrons was measured as a function of the scattering angle $(\phi)$.
It was observed that for an accelerating voltage of $54 \ V$,the intensity of the scattered electrons shows a distinct peak at a scattering angle of $50^{\circ}$.
This peak corresponds to the constructive interference of electron waves scattered from the atomic planes of the crystal.
Therefore,the correct curve is the one that shows a peak at $50^{\circ}$.

(D) The Davisson-Germer experiment provided the first experimental evidence for the wave nature of electrons by demonstrating electron diffraction,which confirmed the de Broglie hypothesis.
Interference and diffraction are characteristic properties of waves.
Since electrons exhibit wave-like behavior,they can indeed undergo interference and diffraction.
Therefore,Statement $1$ is true and Statement $2$ is true,and Statement $2$ provides the physical basis (the property of waves) that explains why the Davisson-Germer experiment was able to demonstrate the wave nature of electrons.

(D) The wave nature of the electron was experimentally confirmed by the Davisson-Germer experiment. In this experiment,a beam of electrons was directed at a nickel crystal,and the resulting diffraction pattern demonstrated that electrons exhibit wave-like properties,such as interference and diffraction,similar to light waves. Therefore,the correct reason is that electrons undergo diffraction by a crystal.

(A) In the Davisson and Germer experiment,the electron gun accelerates electrons using an electric potential $V$.
The kinetic energy $K$ gained by an electron of charge $e$ and mass $m$ is given by $K = eV = \frac{1}{2}mv^2$.
Solving for velocity $v$,we get $v = \sqrt{\frac{2eV}{m}}$.
From this relation,it is clear that the velocity $v$ is directly proportional to the square root of the accelerating potential difference $V$.
Therefore,the velocity of the electrons can be increased by increasing the potential difference between the anode and the filament.

(D) According to the Heisenberg uncertainty principle,the relationship between energy uncertainty $\Delta E$ and time uncertainty $\Delta t$ is given by $\Delta E \cdot \Delta t \geq \frac{h}{4\pi}$.
In Bohr's theory,the energy levels of an electron in an orbit are considered to be perfectly definite,meaning the uncertainty in energy $\Delta E = 0$.
Substituting $\Delta E = 0$ into the uncertainty relation: $0 \cdot \Delta t \geq \frac{h}{4\pi}$,which implies $\Delta t \rightarrow \infty$.
Therefore,the lifetime of an excited state in the context of Bohr's model is infinite.

(C) The Davisson-Germer experiment provided experimental evidence for the de Broglie hypothesis,which states that particles like electrons exhibit wave-like properties.
In the experiment,electrons were scattered by a nickel crystal,and the resulting diffraction pattern confirmed that electrons behave as waves.
Since interference and diffraction are characteristic properties of waves,the wave nature of electrons implies that they must exhibit these phenomena.
Therefore,Statement-$1$ is true,Statement-$2$ is true,and Statement-$2$ provides the theoretical basis (the physical consequence of wave nature) that explains why the experiment was successful in demonstrating the wave nature of electrons.

(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}} \ \mathring{A}$.
Given the accelerating potential $V = 54 \ V$.
Substituting the value into the formula:
$\lambda = \frac{12.27}{\sqrt{54}} \ \mathring{A}$.
$\lambda = \frac{12.27}{7.348} \ \mathring{A}$.
$\lambda \approx 1.67 \ \mathring{A}$.
Thus,the wavelength of the electrons is $1.67 \ \mathring{A}$.

(B) The de Broglie wavelength $\lambda$ of an electron with energy $E$ is given by $\lambda = \frac{12.27}{\sqrt{E}} \mathring{A}$.
For $E = 75 \ eV$,$\lambda = \frac{12.27}{\sqrt{75}} \approx \frac{12.27}{8.66} \approx 1.414 \mathring{A} = \sqrt{2} \mathring{A}$.
In the Davisson-Germer experiment,the scattering angle $\phi$ is the angle between the incident beam and the scattered beam. Given $\phi = 45^o$,the glancing angle $\theta$ with the crystal planes is $\theta = 90^o - \frac{\phi}{2} = 90^o - 22.5^o = 67.5^o$. However,using the standard Bragg's law formula $d \sin \theta = n \lambda$ where $\theta$ is the angle with the atomic planes,and the geometry of the experiment where the scattering angle is $50^o$ (standard) or $45^o$ (given),we use the relation $d \sin(90^o - \phi/2) = n \lambda$.
Given $\phi = 45^o$,$\theta = 90^o - 22.5^o = 67.5^o$.
Using $d \sin(67.5^o) = 1 \times \sqrt{2}$,$d = \frac{1.414}{0.9238} \approx 1.53 \mathring{A}$.
Re-evaluating the provided solution logic: The problem implies the Bragg condition $d \sin \theta = n \lambda$ where $\theta$ is the angle of incidence relative to the planes. If the angle relative to the incident beam is $45^o$,then $\theta = 90^o - 45^o/2 = 67.5^o$. If the question assumes the angle $\theta$ is $45^o$ directly,then $d \sin(45^o) = \sqrt{2} \implies d = \sqrt{2} / (1/\sqrt{2}) = 2 \mathring{A}$.

(C) According to the Heisenberg Uncertainty Principle,the uncertainty in the position of the electron in the $y-$ direction is $\Delta y \simeq d$.
The uncertainty in the momentum in the $y-$ direction is $\Delta P_y \simeq |P_y|$.
From the uncertainty principle,we have $\Delta y \cdot \Delta P_y \simeq h$.
Substituting the values,we get $d \cdot |P_y| \simeq h$.
For the majority of electrons,the momentum acquired is of the order of the uncertainty,thus $|P_y|d \simeq h$.

(B) $(1)$ Davisson and Germer experiment demonstrated the wave nature of electrons through electron diffraction.
$(2)$ Millikan's oil drop experiment was used to determine the elementary charge of an electron.
$(3)$ Rutherford's alpha-particle scattering experiment provided evidence for the existence of a small, dense, positively charged nucleus at the center of the atom.
$(4)$ Franck-Hertz experiment provided experimental evidence for the quantisation of energy levels in atoms.
Therefore, the correct matching is $(1)-(i), (2)-(ii), (3)-(iv), (4)-(iii)$.

(B) The wave nature of matter was proposed by de Broglie.
Davisson and Germer performed an experiment to experimentally verify the wave nature of matter.
This experiment demonstrated that electrons exhibit diffraction patterns similar to light waves when scattered by a crystal lattice.
Therefore,electron diffraction is the phenomenon that best supports the theory that matter has a wave nature.

(A) According to the Heisenberg uncertainty principle,the uncertainty in energy $\Delta E$ and the uncertainty in time $\Delta t$ are related by the expression: $\Delta E \cdot \Delta t \approx \frac{h}{2\pi}$.
Given,the average lifetime of the excited state is $\Delta t = 10^{-8} \, s$.
Substituting the values,we get $\Delta E = \frac{h}{2\pi \cdot \Delta t}$.
Using $h = 6.626 \times 10^{-34} \, J \cdot s$,we have $\Delta E = \frac{6.626 \times 10^{-34}}{2 \times 3.14159 \times 10^{-8}} \approx 1.054 \times 10^{-26} \, J$.
To convert this energy into electron-volts $(eV)$,we divide by the charge of an electron $(1.6 \times 10^{-19} \, C)$:
$\Delta E = \frac{1.054 \times 10^{-26}}{1.6 \times 10^{-19}} \approx 6.59 \times 10^{-8} \, eV$.
Rounding to the nearest provided option,the uncertainty in energy is $6.56 \times 10^{-8} \, eV$.

(B) The Davisson-Germer experiment demonstrated the diffraction of electrons by a crystal lattice.
Since diffraction is a characteristic property of waves,this experiment provided direct experimental evidence for the wave nature of electrons,as proposed by the de Broglie hypothesis.
Therefore,the correct option is $B$.

(B) The condition for constructive interference in crystal diffraction (Bragg's Law) is given by $2d \sin \theta = n \lambda$, where $\theta$ is the glancing angle.
From the figure, the angle of incidence $i$ is measured with respect to the normal. The glancing angle $\theta$ is the angle between the incident ray and the crystal plane, so $\theta = 90^{\circ} - i$.
Given $i = 30^{\circ}$, we have $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
For first-order diffraction, $n = 1$. Substituting the values:
$2 \times (1 \times 10^{-10}\; m) \times \sin(60^{\circ}) = 1 \times \lambda$
$\lambda = 2 \times 10^{-10} \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^{-10}\; m \approx 1.732 \times 10^{-10}\; m$.
The de-Broglie wavelength of an electron accelerated by potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.
Squaring both sides:
$\lambda^2 = \frac{h^2}{2meV} \Rightarrow V = \frac{h^2}{2me\lambda^2}$.
Substituting the values:
$V = \frac{(6.6 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times (\sqrt{3} \times 10^{-10})^2}$
$V = \frac{43.56 \times 10^{-68}}{29.12 \times 10^{-50} \times 3} = \frac{43.56 \times 10^{-68}}{87.36 \times 10^{-50}} \approx 0.4986 \times 10^{18} \times 10^{-68} \times 10^{50} \approx 49.86\; V$.
Thus, $V$ should be about $50\; V$.

(D) For constructive interference,the path difference between waves reflected from successive atomic planes must be an integral multiple of the wavelength,which is given by Bragg's Law: $2d \sin \theta = n \lambda_{dB}$,where $\theta$ is the glancing angle (the angle between the incident ray and the crystal plane).
From the provided figure,the angle $i$ is the angle between the incident ray and the normal to the crystal plane. Therefore,the glancing angle $\theta$ is given by $\theta = 90^{\circ} - i$.
Substituting this into Bragg's Law:
$2d \sin(90^{\circ} - i) = n \lambda_{dB}$
Since $\sin(90^{\circ} - i) = \cos i$,we get:
$2d \cos i = n \lambda_{dB}$

(N/A) Heisenberg's uncertainty principle states that it is impossible to determine simultaneously both the exact position $(\Delta x)$ and the exact momentum $(\Delta p)$ of a microscopic particle.
The mathematical expression for this principle is given by:
$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$
Where:
- $\Delta x$ is the uncertainty in position.
- $\Delta p$ is the uncertainty in momentum.
- $h$ is Planck's constant.
Key implications:
$1$. If the position of a particle is measured precisely $(\Delta x \rightarrow 0)$,the uncertainty in its momentum becomes infinite $(\Delta p \rightarrow \infty)$.
$2$. Conversely,if the momentum is measured precisely $(\Delta p \rightarrow 0)$,the uncertainty in its position becomes infinite $(\Delta x \rightarrow \infty)$.
$3$. This principle implies that for microscopic particles like electrons,we cannot define a precise trajectory,as it is impossible to know both position and momentum at the same time.
$4$. For an electron with a definite momentum $p$,the de Broglie wavelength is $\lambda = \frac{h}{p}$. $A$ wave of a single,definite wavelength extends throughout space,meaning the electron is not localized in any finite region,which is consistent with $\Delta x \rightarrow \infty$.

(N/A) resultant wave obtained by the superposition of waves of different wavelengths is called a wave packet.
The matter wave associated with an electron is not extended in infinite space. It is a wave packet extended about a central wavelength.
In this condition,$\Delta x$ is not infinite but has a definite value which depends on the range of the wave packet.
Also,a wave packet does not have a definite wavelength but is made up of a number of wavelengths around a central wavelength.
With the description of a wave packet using the de-Broglie relation and Born's probability interpretation,one can reproduce Heisenberg's uncertainty principle exactly.
Figure $(a)$ shows a schematic diagram of a wave packet,and figure $(b)$ shows an extended wave with a fixed wavelength.

(N/A) Heisenberg's uncertainty principle states that it is impossible to determine simultaneously both the exact position and the exact momentum of a moving microscopic particle (like an electron) with absolute precision.
Mathematically,it is expressed as:
$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$
Where:
$\Delta x$ is the uncertainty in position,
$\Delta p$ is the uncertainty in momentum,
$h$ is Planck's constant.

(N/A) The wave nature of electrons was first experimentally verified by $C.J. Davisson$ and $L.H. Germer$ in $1927$ and $G.P. Thomson$ in $1928$.
Davisson and Thomson shared the Nobel Prize in $1937$ for their experimental discovery of the diffraction of electrons by crystals.
Experimental arrangement: The figure shows the schematic arrangement of the Davisson and Germer experiment.
It consists of an electron gun which comprises a tungsten filament $F$,coated with barium oxide and heated by a low voltage power supply ($L.T.$ battery). The filament is heated by passing a current through the $L.T.$ battery.
When the filament is heated,electrons are emitted due to thermionic emission,which are then accelerated by an $H.T.$ battery to a desired velocity.
These electrons are made to pass through a cylinder with fine holes along its axis,producing a fine collimated beam. The beam is made to fall on the surface of the nickel crystal.
Electrons are scattered in all directions by the atoms of the crystal.
The intensity of the electron beam scattered in a given direction is measured by an electron detector (collector).
The detector can be moved on a circular scale and is connected to a sensitive galvanometer which records the current.
The deflection of the galvanometer is proportional to the intensity of the electron beam entering the collector.
The entire apparatus is enclosed in an evacuated chamber.

(N/A) In the Davisson and Germer experiment,a detector is moved along a circular scale to measure the intensity of scattered electrons at different positions.
The angle between the incident beam and the scattered beam is called the angle of scattering $(\theta)$.
The intensity of the scattered electron beam is measured for different accelerating voltages $(V)$ and different scattering angles $(\theta)$.
The experiment was performed by varying the accelerating voltage from $44 \ V$ to $68 \ V$.
It was observed that a strong peak appeared in the intensity $(I)$ of the scattered electrons at an accelerating voltage of $54 \ V$ and a scattering angle of $\theta = 50^{\circ}$. This indicates that the number of scattered electrons is maximum at this point.
Let stationary electrons be accelerated by a potential $V$. Their kinetic energy is given by:
$K = eV$
$\therefore \frac{1}{2}mv^2 = eV$
$\therefore \frac{1}{2} \frac{m^2v^2}{m} = eV$
$\therefore \frac{p^2}{2m} = eV$ (where $p = mv$ is momentum)
$\therefore p = \sqrt{2meV}$
The wavelength of the matter wave (de Broglie wavelength) is:
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$
Substituting the values of Planck's constant $(h)$,mass $(m)$,and charge $(e)$,we get:
$\lambda = \frac{1.227}{\sqrt{V}} \text{ nm}$

(A) The wave nature of the electron was proposed by Louis de Broglie in $1924$.
However,the experimental verification of this wave nature was first provided by Clinton Davisson and Lester Germer in $1927$ through their electron diffraction experiment.
They observed that electrons scattered by a nickel crystal show a diffraction pattern similar to $X$-rays,which confirms the wave nature of electrons.

(A) In the Davisson-Germer experiment,the galvanometer is used to measure the intensity of the scattered electron beam at different angles.
The deflection of the galvanometer needle is directly proportional to the number of electrons reaching the detector per unit time,which represents the intensity of the scattered electrons.
Therefore,the deflection is proportional to the intensity of the scattered electrons.

(B) The Davisson-Germer experiment provided experimental evidence for the wave nature of electrons by demonstrating electron diffraction.
In this experiment,electrons emitted from an electron gun are accelerated by a potential difference $V$.
These electrons are then directed towards a nickel crystal.
The intensity of the scattered electron beam is measured as a function of the scattering angle.
It was observed that for an accelerating voltage of $54 \ V$,the intensity of the scattered electron beam reached a maximum at a scattering angle of $50^\circ$.
This result confirmed the de Broglie wavelength hypothesis,as the calculated wavelength matched the experimental value.

(N/A) According to Heisenberg's uncertainty principle,$(\Delta x)(\Delta p) \approx \frac{h}{4\pi}$.
Using the relation $\Delta x \Delta p \approx \frac{h}{2\pi}$ for a confined particle:
$\Delta p = \frac{h}{2\pi \Delta x} = \frac{6.626 \times 10^{-34}}{2 \times 3.1416 \times 10^{-9}} \approx 1.055 \times 10^{-25} \, kg \cdot m/s$.
Given $\Delta p = p$,the momentum $p = 1.055 \times 10^{-25} \, kg \cdot m/s$.
The kinetic energy $E$ is given by $E = \frac{p^2}{2m}$.
$E = \frac{(1.055 \times 10^{-25})^2}{2 \times 9.11 \times 10^{-31}} = \frac{1.113 \times 10^{-50}}{1.822 \times 10^{-30}} \approx 6.109 \times 10^{-21} \, J$.
To convert to $eV$,divide by $1.602 \times 10^{-19} \, J/eV$:
$E = \frac{6.109 \times 10^{-21}}{1.602 \times 10^{-19}} \approx 0.0381 \, eV$.

(A) The wave nature of electrons was proposed by Louis de Broglie in $1924$.
However,the first experimental verification of the wave nature of electrons was provided by Clinton Davisson and Lester Germer in $1927$ through their electron diffraction experiment.
Therefore,Davisson and Germer were the first to verify this property.

(A) The wave nature of electrons was experimentally verified by the Davisson-Germer experiment in $1927$.
This experiment demonstrated electron diffraction,which provided direct evidence for the wave-like behavior of electrons as proposed by Louis de Broglie.

(C) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.
Since $\Delta p = m \Delta v$,we have $\Delta x \propto \frac{1}{m \Delta v}$.
For an ideal gas,the root mean square velocity is $v_{rms} = \sqrt{\frac{3kT}{m}}$,so $\Delta v \propto \frac{1}{\sqrt{m}}$.
Substituting this into the uncertainty relation: $\Delta x \propto \frac{1}{m \cdot (1/\sqrt{m})} = \frac{1}{\sqrt{m}}$.
Therefore,the ratio of uncertainty in position is $\frac{\Delta x_e}{\Delta x_p} = \frac{1/\sqrt{m_e}}{1/\sqrt{m_p}} = \sqrt{\frac{m_p}{m_e}}$.

(A) The Davisson-Germer experiment provided the first experimental evidence for the wave nature of electrons by observing electron diffraction patterns from a nickel crystal.
Since electrons exhibit wave-like properties,they must obey the principles of wave mechanics,which include interference and diffraction.
Therefore,Statement $I$ is true because the experiment confirms the wave nature of electrons.
Statement $II$ is also true because diffraction and interference are characteristic properties of waves,and electrons demonstrate these behaviors due to their wave nature.
Thus,both statements are correct.

(B) The Heisenberg uncertainty principle states that $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$. The lowest energy of a particle confined in a region of size $\Delta x$ is proportional to $\frac{1}{m(\Delta x)^2}$.
$(IV)$ $A$ $H_{2}$ molecule in a nanotube is confined to a relatively large space (nanometer scale) and has a large mass,leading to the lowest energy.
$(II)$ $A$ hydrogen atom in a $H_{2}$ molecule is confined by interatomic forces within the molecule,resulting in a smaller confinement region than the nanotube,thus higher energy than $(IV)$.
$(I)$ An electron in a $H_{2}$ molecule is much lighter than an atom,and its confinement within the molecular orbital leads to a higher kinetic energy than the atom.
$(III)$ $A$ proton in the carbon nucleus is confined to an extremely small region (femtometer scale) by strong nuclear forces,resulting in the highest energy due to the uncertainty principle.
Therefore,the correct order of increasing lowest energy is $IV < II < I < III$.

(D) $1$. Assertion $A$ states that a beam of electrons exhibits wave nature,including interference and diffraction. This is a fundamental property of matter waves as proposed by de Broglie.
$2$. Reason $R$ states that the Davisson-Germer experiment experimentally verified the wave nature of electrons. This is a historical fact.
$3$. The wave nature of electrons (Assertion $A$) is confirmed by the diffraction patterns observed in the Davisson-Germer experiment (Reason $R$). Therefore,the experimental verification provides the physical evidence that explains why we conclude electrons exhibit wave-like properties such as interference and diffraction.
$4$. Thus,both statements are correct,and $R$ is the correct explanation of $A$.

(A) $G$ $P$ Thomson experimentally confirmed the existence of matter waves (de-Broglie's hypothesis) by demonstrating that electron beams undergo diffraction when they are scattered by the regular atomic arrays of crystals. This experiment provided direct evidence for the wave nature of matter.