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Unit4 - Subjective Questions

PHY110 • Practice Questions with Detailed Answers

Discuss the failure of Classical Mechanics and the need for Quantum Mechanics.

Classical mechanics, based on Newton's laws and Maxwell's electromagnetic theory, successfully explains macroscopic phenomena (motion of planets, projectiles, etc.). However, it fails to explain microscopic phenomena.

Main failures leading to the need for Quantum Mechanics:

Black Body Radiation: Classical theory (Rayleigh-Jeans law) predicted that at short wavelengths, the energy emitted by a black body goes to infinity (Ultraviolet Catastrophe), which contradicted experimental results.
Photoelectric Effect: Classical wave theory could not explain why light below a certain threshold frequency fails to eject electrons, regardless of intensity, or why emission is instantaneous.
Atomic Stability: According to classical electrodynamics, an accelerating electron in an orbit should radiate energy and spiral into the nucleus, making atoms unstable. Quantum mechanics explains atomic stability via discrete energy levels.
Heat Capacity of Solids: Classical variation of specific heat with temperature failed at low temperatures.

Explain the Photoelectric effect and state the laws of photoelectric emission.

Photoelectric Effect:
The phenomenon of emission of electrons from a metal surface when electromagnetic radiation (light) of suitable frequency is incident on it is called the photoelectric effect.

Laws of Photoelectric Emission:

Threshold Frequency: For every metal, there exists a minimum frequency of incident light ($
u_0$) below which no photo-emission takes place.
Instantaneous Process: The emission of photoelectrons is an instantaneous process (time lag $< 10^{-9}$ s).
Intensity Dependence: The rate of emission of photoelectrons (saturation current) is directly proportional to the intensity of incident light.
Kinetic Energy: The maximum kinetic energy of the emitted photoelectrons is directly proportional to the frequency of incident light and is independent of intensity.

Derive Einstein's Photoelectric equation.

Einstein applied Planck's quantum theory to the photoelectric effect. Light consists of bundles of energy called photons with energy $E = h
u$.

When a photon of energy $h
u$ strikes a metal surface, its energy is used in two ways:

Work Function ( $\Phi$ or $W$ ): A part of the energy is used to overcome the surface barrier to eject the electron. $\Phi = h
u_0$.
Kinetic Energy ( $K_{max}$ ): The remaining energy is imparted to the electron as maximum kinetic energy.

Equation:
By conservation of energy:

Energy_{incident} = Work\ Function + Max\ Kinetic\ Energy

h u = \Phi + \frac{1}{2}mv_{max}^2

h u = h u_0 + K_{max}

K_{max} = h( u - u_0)

This is Einstein's Photoelectric equation.

State de Broglie's hypothesis and derive the expression for the wavelength of matter waves.

de Broglie Hypothesis:
Louis de Broglie suggested that nature loves symmetry. If radiation (light) has a dual nature (wave and particle), then matter (particles like electrons, protons) must also possess a dual nature. A moving particle carries a wave associated with it, called a matter wave or pilot wave.

Derivation:

From Planck's theory, energy of a photon: $E = h
u = \frac{hc}{\lambda}$.
From Einstein's mass-energy relation: $E = mc^2$ .

Equating both:

mc^2 = \frac{hc}{\lambda}

\lambda = \frac{h}{mc}

For a material particle of mass $m$ moving with velocity $v$ , we replace $c$ with $v$ :

\lambda = \frac{h}{mv}

Since momentum

p = mv

\lambda = \frac{h}{p}

This is the de Broglie wavelength expression.

Derive the expression for the de Broglie wavelength of an electron accelerated by a potential difference $V$ .

Consider an electron of mass $m$ and charge $e$ accelerated from rest through a potential difference $V$ .

Kinetic Energy:
The work done on the electron is stored as kinetic energy ( $E$ ):

E = eV

Also, Kinetic Energy is given by:

E = \frac{1}{2}mv^2 = \frac{p^2}{2m}

Momentum:

p^2 = 2mE \implies p = \sqrt{2mE}

Substituting

E = eV

p = \sqrt{2meV}

Wavelength:
According to de Broglie's equation $\lambda = \frac{h}{p}$ :

\lambda = \frac{h}{\sqrt{2meV}}

Numerical Value:
Substituting standard values ( $h = 6.626 \times 10^{-34}$ Js, $m = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C):

\lambda = \frac{12.27}{\sqrt{V}} \ \mathring{A}

Write the expressions for de Broglie wavelength in terms of Kinetic Energy and Temperature for gas molecules.

1. In terms of Kinetic Energy ( $E$ ):
We know $E = \frac{p^2}{2m}$ , so $p = \sqrt{2mE}$ .
Therefore,

\lambda = \frac{h}{\sqrt{2mE}}

2. In terms of Temperature ( $T$ ):
For a gas molecule at absolute temperature $T$ , the average kinetic energy according to kinetic theory of gases is:

E = \frac{3}{2}k_BT

where

k_B

is the Boltzmann constant.

Substituting this into the momentum equation:

p = \sqrt{2m \left(\frac{3}{2}k_BT\right)} = \sqrt{3mk_BT}

Therefore, the wavelength is:

\lambda = \frac{h}{\sqrt{3mk_BT}}

State Heisenberg's Uncertainty Principle and discuss its physical significance.

Statement:
It is impossible to determine simultaneously both the exact position and the exact momentum of a moving particle with infinite accuracy. The product of the uncertainties in position ( $\Delta x$ ) and momentum ( $\Delta p_x$ ) is always greater than or equal to $\frac{h}{4\pi}$ (or $\frac{\hbar}{2}$ ).

Mathematical Form:

\Delta x \cdot \Delta p_x \geq \frac{\hbar}{2}

Similarly for Energy and Time:

\Delta E \cdot \Delta t \geq \frac{\hbar}{2}

Physical Significance:

Breakdown of Bohr's Model: It implies that electrons cannot move in well-defined fixed orbits, as that would mean definite position and momentum.
Probabilistic Nature: It introduces the concept of probability in finding a particle, leading to the development of wave mechanics.
Macroscopic vs Microscopic: The principle is negligible for macroscopic objects (due to large mass) but significant for microscopic particles like electrons.

Show that electrons cannot exist inside the nucleus using the Heisenberg Uncertainty Principle.

Assume an electron exists inside the nucleus.

Uncertainty in Position: The diameter of a typical nucleus is approx $10^{-14}$ m. Thus, the maximum uncertainty in the position of the electron is:

$\Delta x \approx 10^{-14}\ \text{m}$
Uncertainty in Momentum: Using $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$ :

$\Delta p \geq \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 10^{-14}} \approx 5.28 \times 10^{-21}\ \text{kg m/s}$
Minimum Energy: The momentum $p$ must be at least comparable to $\Delta p$ . The energy $E$ can be calculated using $E = pc$ (relativistic approximation for high speed):

$E \approx (5.28 \times 10^{-21}) \times (3 \times 10^8)\ \text{Joules}$
$E \approx 1.58 \times 10^{-12}\ \text{J}$
Converting to MeV ( $1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J}$ ):
$E \approx \frac{1.58 \times 10^{-12}}{1.6 \times 10^{-13}} \approx 9.9\ \text{MeV} \approx 10\ \text{MeV}$

Conclusion: Experimental evidence (Beta decay) shows electrons emitted from nuclei have energies of only 2-3 MeV. Since the required confinement energy (~10 MeV) is much higher, electrons cannot exist inside the nucleus.

Differentiate between Phase Velocity and Group Velocity.

Phase Velocity ( $v_p$ ):

Definition: The velocity with which a single monochromatic wave (a definite phase) travels through a medium.
Formula: $v_p = \frac{\omega}{k} = \frac{E}{p}$ .
Significance: It represents the speed of the phase of the wave. For matter waves, $v_p$ is often calculated as $c^2/v$ , which exceeds the speed of light ( $c$ ), having no physical meaning for carrying information.

Group Velocity ( $v_g$ ):

Definition: The velocity with which a 'wave packet' (group of waves of slightly different frequencies) travels.
Formula: $v_g = \frac{d\omega}{dk} = \frac{dE}{dp}$ .
Significance: It represents the velocity of the particle (energy envelope). For a material particle, the group velocity is equal to the particle velocity ( $v$ ).

Derive the relation between Phase Velocity ( $v_p$ ) and Group Velocity ( $v_g$ ).

We know:

v_p = \frac{\omega}{k} \implies \omega = k \cdot v_p

And:

v_g = \frac{d\omega}{dk}

Substitute $\omega$ in the group velocity equation:

v_g = \frac{d(k v_p)}{dk}

v_g = v_p + k \frac{dv_p}{dk}

We know $k = \frac{2\pi}{\lambda}$ , so differentiating $k$ with respect to $\lambda$ :

\frac{dk}{d\lambda} = -\frac{2\pi}{\lambda^2} = -\frac{k}{\lambda}

Using chain rule:

\frac{dv_p}{dk} = \frac{dv_p}{d\lambda} \cdot \frac{d\lambda}{dk} = \frac{dv_p}{d\lambda} \cdot \left(-\frac{\lambda}{k}\right)

Substituting back:

v_g = v_p + k \left( -\frac{\lambda}{k} \frac{dv_p}{d\lambda} \right)

v_g = v_p - \lambda \frac{dv_p}{d\lambda}

Conclusion:

If medium is non-dispersive ( $\frac{dv_p}{d\lambda} = 0$ ), then $v_g = v_p$ .
If medium is dispersive ( $\frac{dv_p}{d\lambda} > 0$ ), then $v_g < v_p$ .

Prove that the group velocity of matter waves associated with a moving particle is equal to the particle's velocity.

Consider a particle of mass $m$ moving with velocity $v$ .
According to relativistic energy theory:

E = mc^2 = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}}

And angular frequency

\omega = \frac{E}{\hbar}

Momentum $p = mv = \frac{m_0 v}{\sqrt{1 - v^2/c^2}}$
And propagation constant $k = \frac{p}{\hbar}$ .

Group velocity is given by:

v_g = \frac{d\omega}{dk} = \frac{dE/\hbar}{dp/\hbar} = \frac{dE}{dp}

From $E^2 = p^2c^2 + m_0^2c^4$ , differentiate with respect to $p$ :

2E \frac{dE}{dp} = 2pc^2 + 0

\frac{dE}{dp} = \frac{pc^2}{E}

Substitute $p = mv$ and $E = mc^2$ :

v_g = \frac{(mv)c^2}{mc^2}

v_g = v

Thus, the group velocity of the matter wave is equal to the classical velocity of the particle.

Define Wave Function ( $\Psi$ ) and discuss its physical significance and characteristics.

Wave Function ( $\Psi$ ):
It is a mathematical quantity (generally complex) that describes the quantum state of a particle in space and time. It contains all the information about the system.

Physical Significance (Born Interpretation):

The wave function $\Psi$ itself has no direct physical meaning because it can be complex.
The quantity $|\Psi(x,t)|^2 = \Psi \cdot \Psi^*$ represents the Probability Density.
It gives the probability of finding the particle at a particular position $x$ at time $t$ .
The total probability over all space must be 1: $\int_{-\infty}^{\infty} |\Psi|^2 dx = 1$ (Normalization condition).

Characteristics of a Well-Behaved Wave Function:

$\Psi$ must be finite everywhere.
$\Psi$ must be single-valued (one probability for one location).
$\Psi$ and its derivative $\frac{\partial \Psi}{\partial x}$ must be continuous everywhere.

Derive Schrödinger's Time Independent Wave Equation (STIWE).

Starting Point:
Consider a particle of mass $m$ moving along the x-axis. The classical total energy $E$ is sum of kinetic ( $K$ ) and potential ( $V$ ) energy.

E = K + V = \frac{p^2}{2m} + V(x)

Wave Function:
The wave function for a free particle is $\Psi(x) = A e^{i(kx)}$ .
Differentiating twice w.r.t $x$ :

\frac{d\Psi}{dx} = ik\Psi

\frac{d^2\Psi}{dx^2} = (ik)^2\Psi = -k^2\Psi

Since

p = \hbar k

, we have

k = p/\hbar

\frac{d^2\Psi}{dx^2} = -\frac{p^2}{\hbar^2}\Psi \implies p^2\Psi = -\hbar^2 \frac{d^2\Psi}{dx^2}

Operator Substitution:
Multiply the energy equation by $\Psi$ :

E\Psi = \frac{p^2}{2m}\Psi + V\Psi

Substitute the value of

p^2\Psi

E\Psi = \frac{1}{2m} \left( -\hbar^2 \frac{d^2\Psi}{dx^2} \right) + V\Psi

E\Psi = -\frac{\hbar^2}{2m} \frac{d^2\Psi}{dx^2} + V\Psi

Rearranging:

\frac{d^2\Psi}{dx^2} + \frac{2m}{\hbar^2}(E - V)\Psi = 0

This is the one-dimensional Schrödinger Time Independent Wave Equation.

Derive Schrödinger's Time Dependent Wave Equation (STDWE).

Wave Function:
Consider the wave function of a free particle: $\Psi(x,t) = A e^{-i\omega(t - x/v)} = A e^{-i(\omega t - kx)}$ .
Using $E = \hbar \omega$ and $p = \hbar k$ , we get:

\Psi(x,t) = A e^{-\frac{i}{\hbar}(Et - px)}

Energy Operator:
Differentiate $\Psi$ w.r.t time $t$ :

\frac{\partial \Psi}{\partial t} = -\frac{iE}{\hbar} \Psi

E\Psi = -\frac{\hbar}{i} \frac{\partial \Psi}{\partial t} = i\hbar \frac{\partial \Psi}{\partial t}

Momentum Operator:
Differentiate $\Psi$ w.r.t position $x$ twice:

\frac{\partial^2 \Psi}{\partial x^2} = \left(\frac{ip}{\hbar}\right)^2 \Psi = -\frac{p^2}{\hbar^2} \Psi

p^2\Psi = -\hbar^2 \frac{\partial^2 \Psi}{\partial x^2}

Total Energy Equation:
Total Energy $E = \frac{p^2}{2m} + V$ .
Multiply by $\Psi$ :

E\Psi = \frac{p^2}{2m}\Psi + V\Psi

Substitute the operator values:

i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V(x,t)\Psi

This is Schrödinger's Time Dependent Wave Equation.

Apply Schrödinger's wave equation to a particle in a 1D rigid box and derive the expression for Energy Eigenvalues.

System Setup:
Particle of mass $m$ in a box of length $L$ .

$V(x) = 0$ for $0 < x < L$
$V(x) = \infty$ for $x \leq 0$ and $x \geq L$

Equation:
Inside the box ( $V=0$ ), the Schrödinger equation is:

\frac{d^2\Psi}{dx^2} + \frac{2mE}{\hbar^2}\Psi = 0

Let

k^2 = \frac{2mE}{\hbar^2}

. The solution is:

\Psi(x) = A \sin(kx) + B \cos(kx)

Boundary Conditions:

At $x=0, \Psi=0 \implies B=0$ . Thus $\Psi = A \sin(kx)$ .
At $x=L, \Psi=0 \implies A \sin(kL) = 0$ .
Since $A
eq 0 $(particle must exist),$ \sin(kL) = 0$.
$kL = n\pi \quad (n = 1, 2, 3...)$
$k = \frac{n\pi}{L}$

Energy Eigenvalues:
Substitute $k$ back into energy relation:

\frac{2mE}{\hbar^2} = \frac{n^2\pi^2}{L^2}

E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}

Using

\hbar = h/2\pi

E_n = \frac{n^2 h^2}{8mL^2}

These are the discrete energy levels.

Determine the normalized wave functions for a particle in a 1D box of width L.

Wave Function Form:
From the particle in a box derivation, we found:

\Psi_n(x) = A \sin\left(\frac{n\pi x}{L}\right)

Normalization Condition:
The total probability of finding the particle inside the box is 1.

\int_{0}^{L} |\Psi_n(x)|^2 dx = 1

\int_{0}^{L} A^2 \sin^2\left(\frac{n\pi x}{L}\right) dx = 1

A^2 \int_{0}^{L} \frac{1 - \cos(2n\pi x/L)}{2} dx = 1

A^2 \left[ \frac{x}{2} - \frac{\sin(2n\pi x/L)}{4n\pi/L} \right]_0^L = 1

Evaluating limits:

A^2 \left[ \left(\frac{L}{2} - 0\right) - (0 - 0) \right] = 1

A^2 \frac{L}{2} = 1 \implies A = \sqrt{\frac{2}{L}}

Normalized Wave Function:

\Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)

What is Zero Point Energy? Calculate it for a particle in a box.

Definition:
Zero Point Energy is the lowest possible energy that a quantum mechanical system may have. Unlike classical mechanics where the lowest energy can be zero (particle at rest), quantum particles always possess some kinetic energy due to the uncertainty principle.

For Particle in a Box:
The energy eigenvalues are given by:

E_n = \frac{n^2 h^2}{8mL^2}

where

n = 1, 2, 3, ...

Note that $n$ cannot be 0, because if $n=0$ , the wave function $\Psi$ becomes zero everywhere, meaning the particle does not exist in the box.

Therefore, the minimum energy corresponds to $n=1$ :

E_{min} = E_1 = \frac{1^2 h^2}{8mL^2} = \frac{h^2}{8mL^2}

This minimum non-zero energy is the Zero Point Energy.

Explain the concept of Quantum Tunneling (Qualitative idea).

Concept:
Quantum tunneling is a phenomenon where a particle penetrates through a potential energy barrier that represents a height greater than the total energy of the particle.

Classical vs Quantum:

Classically: If a particle encounters a potential barrier $V > E$ (Energy of particle), it will be reflected back completely. It cannot cross the region.
Quantum Mechanically: The wave function $\Psi$ does not become zero immediately at the barrier boundary. Instead, it decays exponentially inside the barrier. If the barrier is thin enough, the wave function has a non-zero value on the other side.

Mechanism:
This implies there is a finite probability ( $|\Psi|^2 > 0$ ) that the particle 'tunnels' through the forbidden region and appears on the other side. This is a direct consequence of the wave nature of matter.

List applications of the Tunneling Effect.

The quantum tunneling effect explains various phenomena and is used in modern devices:

Alpha Decay: Explains how alpha particles escape the nucleus despite the strong nuclear potential barrier which is higher than the energy of the alpha particle.
Scanning Tunneling Microscope (STM): Uses electron tunneling between a sharp tip and a surface to image atoms with high resolution.
Tunnel Diode: A semiconductor diode that utilizes tunneling to achieve negative resistance, used in high-frequency oscillators.
Nuclear Fusion in Stars: Protons tunnel through the Coulomb repulsion barrier to fuse together, powering stars like the Sun.

Compare Matter Waves and Electromagnetic Waves.

Matter Waves:

Associated with moving particles (electrons, protons, etc.).
Wavelength: $\lambda = h/mv$ .
They are not electromagnetic in nature; they do not consist of electric and magnetic fields.
Can travel in a vacuum but require a moving mass.
Velocity ( $v_p$ ) can be greater than the speed of light (phase velocity), but the group velocity is less than $c$ .

Electromagnetic (EM) Waves:

Radiated by accelerated charged particles.
Wavelength: $\lambda = c/
u$.
Consist of oscillating electric and magnetic fields perpendicular to each other.
Can travel through a vacuum.
Travel at a constant speed of light $c$ in a vacuum.

Unit3 Unit5