Set B Solution jj

Below is a step‐by‐step solution to each part of the multi–problem question.

---

1. Normal Zeeman Splitting of the Calcium 4226 Å Line

For the normal Zeeman effect (which applies when the involved level has zero net spin), the energy shift between adjacent Zeeman components is given by

$$\Delta E = \mu_B B,$$

where

• $\mu_B \approx 9.274\times10^{-24}\,\mathrm{J/T}$ is the Bohr magneton, and
• $B = 1.21\,\mathrm{T}$ is the magnetic field.

Thus, the energy splitting is

$$\Delta E \approx 9.274\times10^{-24}\,\mathrm{J/T} \times 1.21\,\mathrm{T} \approx 1.12\times10^{-23}\,\mathrm{J}.$$

To relate this energy shift to a wavelength shift, we use the relation

$$\Delta E = \frac{hc}{\lambda^2}\Delta\lambda,$$

where

• $h=6.626\times10^{-34}\,\mathrm{J\cdot s}$,
• $c=3.00\times10^8\,\mathrm{m/s}$,
• and the wavelength $\lambda = 4226\,\text{Å} = 4.226\times10^{-7}\,\mathrm{m}$.

Solving for $\Delta\lambda$:

$$\Delta\lambda = \frac{\lambda^2}{hc}\Delta E.$$

Calculating step‐by‐step:

1. $\lambda^2 = (4.226\times10^{-7}\,\mathrm{m})^2 \approx 1.787\times10^{-13}\,\mathrm{m^2}$.
2. $hc \approx 6.626\times10^{-34}\,\mathrm{J\cdot s} \times 3.00\times10^8\,\mathrm{m/s} \approx 1.9878\times10^{-25}\,\mathrm{J\cdot m}$.
3. Then, $$\Delta\lambda \approx \frac{1.787\times10^{-13}\,\mathrm{m^2}}{1.9878\times10^{-25}\,\mathrm{J\cdot m}} \times 1.12\times10^{-23}\,\mathrm{J} \approx 1.01\times10^{-11}\,\mathrm{m}.$$

In other words, the Zeeman splitting corresponds to an energy difference of about $1.12\times10^{-23}\,\mathrm{J}$ and a wavelength shift of approximately $1.01\times10^{-11}\,\mathrm{m}$ (or 0.0101 Å).

---

2. The Hall Effect and Hall Voltage

Definition:
The Hall effect occurs when a current–carrying conductor or semiconductor is placed in a magnetic field that is perpendicular to the current. The magnetic Lorentz force deflects the moving charge carriers (electrons or holes) to one side of the material. This deflection causes a buildup of charge and produces a transverse potential difference called the Hall voltage.

Hall Voltage:
The Hall voltage $V_H$ is given by

$$V_H = R_H \frac{IB}{d},$$

where

• $I$ is the current,
• $B$ is the magnetic field,
• $d$ is the thickness of the conductor,
• and the Hall coefficient $R_H$ is defined as

$$R_H = \frac{1}{nq},$$

with $n$ being the charge carrier density and $q$ the charge of the carriers.

Determining the Nature of Charge Carriers:
The sign of the Hall coefficient $R_H$ reveals the type of charge carrier:

• A negative $R_H$ indicates that the carriers are negatively charged (electrons).
• A positive $R_H$ indicates that the carriers are positively charged (holes).

Thus, by measuring the Hall voltage and knowing the current, magnetic field, and sample dimensions, one can determine both the density and the sign of the charge carriers in a material.

---

3. Free Particle in an Infinite Potential Well

Consider a one-dimensional infinite potential well (particle in a box) of width $L$:

• The potential is $V(x)=0$ for $0<x<L$ and $V(x)=\infty$ outside this interval.

The allowed energy levels are given by

$$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2},\quad n=1,2,3,\dots$$

and the corresponding normalized wave functions are

$$\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right),\quad 0<x<L,$$

with $\psi(x)=0$ elsewhere.

---

4. Fermi Energy of Copper

For a free electron gas, the Fermi energy $E_F$ is given by

$$E_F = \frac{\hbar^2}{2m_e}(3\pi^2 n)^{2/3},$$

where

• $n = 8.4\times10^{28}\,\mathrm{m}^{-3}$ is the electron density,
• $\hbar = 1.055\times10^{-34}\,\mathrm{J\cdot s}$,
• and $m_e = 9.11\times10^{-31}\,\mathrm{kg}$.

Step 1. Calculate the factor:

$$3\pi^2 n \approx 3\pi^2 \times 8.4\times10^{28}\,\mathrm{m}^{-3} \approx 2.49\times10^{30}\,\mathrm{m}^{-3}.$$

Step 2. Raise to the $\frac{2}{3}$ power:

$$(3\pi^2 n)^{2/3} \approx 1.84\times10^{20}\,\mathrm{m}^{-2}.$$

Step 3. Compute $E_F$:

$$E_F \approx \frac{(1.055\times10^{-34})^2}{2\times9.11\times10^{-31}} \times 1.84\times10^{20} \approx 1.12\times10^{-18}\,\mathrm{J}.$$

Converting to electron volts (using $1\,\mathrm{eV} \approx 1.602\times10^{-19}\,\mathrm{J}$):

$$E_F \approx \frac{1.12\times10^{-18}}{1.602\times10^{-19}} \approx 7.0\,\mathrm{eV}.$$

(b) Temperature Comparison:
To find the temperature $T$ at which the average thermal energy $kT$ equals the Fermi energy, set

$$kT = E_F.$$

Using $k=1.38\times10^{-23}\,\mathrm{J/K}$:

$$T \approx \frac{1.12\times10^{-18}\,\mathrm{J}}{1.38\times10^{-23}\,\mathrm{J/K}} \approx 8.12\times10^4\,\mathrm{K}.$$

Thus, $kT \approx E_F$ at about $8.1\times10^4\,\mathrm{K}$.

---

5. The Classical Free Electron Model

The classical free electron model (also known as the Drude model) makes the following assumptions:

• Electrons as Classical Particles: Conduction electrons in a metal are treated as a gas of free, non–interacting particles.
• Random Motion and Collisions: Electrons move randomly and occasionally collide with the fixed ions in the lattice. Collisions are characterized by an average time between collisions, called the mean free time $\tau$.
• Ohm’s Law: The model explains electrical conductivity by relating the drift velocity acquired by electrons under an applied electric field to $\tau$ and the density of free electrons.
• Limitations: While it successfully predicts some macroscopic properties such as electrical and thermal conductivity, the model cannot explain all observed phenomena (for example, it fails to account for quantum effects and the temperature dependence of the electronic heat capacity).

---

6. Dynamics of a Rotating Wheel with a Descending Mass

A wheel of radius $R=0.4\,\mathrm{m}$ and moment of inertia $I=1.2\,\mathrm{kg\cdot m^2}$ has a rope wound around it with a $2\,\mathrm{kg}$ weight attached. The weight descends a distance $h=1.5\,\mathrm{m}$. Assume no friction and use $g=9.8\,\mathrm{m/s^2}$.

Energy Conservation:

The loss in gravitational potential energy is

$$mgh = 2\,\mathrm{kg} \times 9.8\,\mathrm{m/s^2} \times 1.5\,\mathrm{m} = 29.4\,\mathrm{J}.$$

This energy converts into:

• Translational kinetic energy of the mass: $\frac{1}{2}mv^2$,
• Rotational kinetic energy of the wheel: $\frac{1}{2}I\omega^2$.

Since the rope unwinds without slipping, the linear speed $v$ of the mass is related to the angular speed $\omega$ of the wheel by

$$v = \omega R.$$

Thus, the total kinetic energy is

$$\frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{R}\right)^2 = \frac{1}{2}v^2\left(m + \frac{I}{R^2}\right).$$

Setting this equal to the lost potential energy:

$$29.4 = \frac{1}{2}v^2\left(2 + \frac{1.2}{(0.4)^2}\right).$$

Calculate $I/R^2$:

$$\frac{1.2}{(0.4)^2} = \frac{1.2}{0.16} = 7.5.$$

Thus,

$$29.4 = \frac{1}{2}v^2 (2 + 7.5) = \frac{1}{2}v^2 (9.5).$$

Solving for $v$:

$$v^2 = \frac{29.4\times2}{9.5} \approx \frac{58.8}{9.5} \approx 6.19,$$ $$v \approx \sqrt{6.19} \approx 2.49\,\mathrm{m/s}.$$

Then, the rotational (angular) speed is

$$\omega = \frac{v}{R} \approx \frac{2.49}{0.4} \approx 6.23\,\mathrm{rad/s}.$$

---

7. Electric Field from Two Point Charges

We have two positive charges on the x–axis:

• $q_1 = 3\times10^{-9}\,\mathrm{C}$ at $x=0$,
• $q_2 = 5\times10^{-9}\,\mathrm{C}$ at $x=4\,\mathrm{m}$.

(a) Electric Field at the Midpoint ($x=2\,\mathrm{m}$)

For a point charge, the electric field is

$$E=\frac{kq}{r^2},\quad k\approx9\times10^9\,\mathrm{N\cdot m^2/C^2}.$$

• From $q_1$:
  Distance $r_1=2\,\mathrm{m}$

  $$E_1 = \frac{9\times10^9\times3\times10^{-9}}{(2)^2} = \frac{27}{4} \approx 6.75\,\mathrm{N/C}.$$

  The field points to the right (away from $q_1$).

• From $q_2$:
  Distance $r_2=4-2=2\,\mathrm{m}$

  $$E_2 = \frac{9\times10^9\times5\times10^{-9}}{(2)^2} = \frac{45}{4} \approx 11.25\,\mathrm{N/C}.$$

  Here the field points to the left (away from $q_2$, since the point is to its left).

The net electric field at $x=2\,\mathrm{m}$ is

$$E_{\text{net}} = E_1 - E_2 \approx 6.75 - 11.25 = -4.5\,\mathrm{N/C}.$$

The negative sign indicates that the net field is $4.5\,\mathrm{N/C}$ directed to the left.

(b) Location Where the Resultant Field is Zero

Let $x$ be the point (with $0<x<4$) where the net field vanishes. The field due to $q_1$ (at $x=0$) at a point $x$ is

$$E_1=\frac{kq_1}{x^2}$$

(directed to the right) and the field due to $q_2$ (at $x=4\,\mathrm{m}$) is

$$E_2=\frac{kq_2}{(4-x)^2}$$

(directed to the left). Setting the magnitudes equal for cancellation:

$$\frac{q_1}{x^2} = \frac{q_2}{(4-x)^2}.$$

Plug in $q_1=3\times10^{-9}\,\mathrm{C}$ and $q_2=5\times10^{-9}\,\mathrm{C}$:

$$\frac{3}{x^2} = \frac{5}{(4-x)^2} \quad \Longrightarrow \quad \left(\frac{4-x}{x}\right)^2 = \frac{5}{3}.$$

Taking the square root,

$$\frac{4-x}{x} = \sqrt{\frac{5}{3}} \approx 1.29.$$

Solve for $x$:

$$4-x = 1.29\,x \quad \Longrightarrow \quad 4 = 2.29\,x,$$ $$x \approx \frac{4}{2.29} \approx 1.75\,\mathrm{m}.$$

Thus, the net electric field is zero at approximately $1.75\,\mathrm{m}$ from the origin along the x–axis.

---

8. Wavelengths of the Paschen Series of Hydrogen

For the Paschen series, electrons drop to the $n_f=3$ level. The Rydberg formula is

$$\frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n^2}\right),\quad n>3,$$

with $R=1.09\times10^7\,\mathrm{m}^{-1}$.

Shortest Wavelength (Series Limit):
For $n\to\infty$,

$$\frac{1}{\lambda_{\text{min}}} = R\left(\frac{1}{9} - 0\right) = \frac{R}{9},$$ $$\lambda_{\text{min}} = \frac{9}{R} = \frac{9}{1.09\times10^7} \approx 8.26\times10^{-7}\,\mathrm{m}\quad (826\,\mathrm{nm}).$$

Longest Wavelength (Transition $n=4 \to n=3$):

$$\frac{1}{\lambda_{\text{max}}} = R\left(\frac{1}{9} - \frac{1}{16}\right) = R\left(\frac{16-9}{144}\right) = \frac{7R}{144},$$ $$\lambda_{\text{max}} = \frac{144}{7R} = \frac{144}{7\times1.09\times10^7} \approx 1.89\times10^{-6}\,\mathrm{m}\quad (1890\,\mathrm{nm}).$$

---

9. Uncertainty in Velocity When $\Delta x = \lambda_{\text{dB}}$

The de Broglie wavelength for a particle of mass $m$ moving with velocity $v$ is

$$\lambda = \frac{h}{mv}.$$

If the uncertainty in position is taken to be $\Delta x = \lambda$, then by the Heisenberg uncertainty principle

$$\Delta x\,\Delta p \ge \frac{\hbar}{2},$$

with $\Delta p = m\,\Delta v$. Thus,

$$\Delta v \ge \frac{\hbar}{2m\,\Delta x}.$$

Substitute $\Delta x = \lambda = \frac{h}{mv}$ and note that $\hbar = \frac{h}{2\pi}$:

$$\Delta v \ge \frac{h/(2\pi)}{2m\,(h/(mv))} = \frac{h}{2\pi} \cdot \frac{mv}{2mh} = \frac{v}{4\pi}.$$

So the uncertainty in the velocity is

$$\Delta v \approx \frac{v}{4\pi}.$$

---

Summary of Answers

1. Zeeman Splitting:

   • Energy splitting: $\Delta E \approx 1.12\times10^{-23}\,\mathrm{J}$.
   • Wavelength shift: $\Delta\lambda \approx 1.01\times10^{-11}\,\mathrm{m}$ (≈0.0101 Å).

2. Hall Effect:

   • Definition: Generation of a transverse voltage (Hall voltage) in a current–carrying conductor placed in a perpendicular magnetic field.
   • Determination: The sign of the Hall coefficient $R_H = 1/(nq)$ shows whether the charge carriers are negative (electrons) or positive (holes).

3. Infinite Potential Well:

   • Energies: $E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$.
   • Wave functions: $\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$.

4. Fermi Energy of Cu:

   • $E_F \approx 1.12\times10^{-18}\,\mathrm{J}$ (≈7.0 eV).
   • Corresponding temperature: $T \approx 8.1\times10^4\,\mathrm{K}$.

5. Classical Free Electron Model:

   • Treats conduction electrons as a classical gas that randomly moves and collides with fixed ions; explains electrical and thermal conductivity via kinetic theory.

6. Rotating Wheel Dynamics:

   • Downward velocity of the weight: $v \approx 2.49\,\mathrm{m/s}$.
   • Angular velocity of the wheel: $\omega \approx 6.23\,\mathrm{rad/s}$.

7. Electric Field from Two Charges:

   • (a) At $x=2\,\mathrm{m}$: Net field $E_{\text{net}} \approx 4.5\,\mathrm{N/C}$ to the left.
   • (b) Field zero at $x \approx 1.75\,\mathrm{m}$ from the origin.

8. Paschen Series Wavelengths:

   • Shortest wavelength (series limit): $\lambda_{\text{min}} \approx 826\,\mathrm{nm}$.
   • Longest wavelength (transition from $n=4$ to $n=3$): $\lambda_{\text{max}} \approx 1890\,\mathrm{nm}$.

9. Uncertainty in Velocity:

   • If $\Delta x = \lambda = \frac{h}{mv}$, then
     $\Delta v \approx \frac{v}{4\pi}$.

Each result has been derived using standard physical relations and assumptions.