110 marksThe semiconductor diodeAnswer

Explain equilibrium current across the PN junction? Use Fermi-Dirac statistics and Maxwell-Boltzmann distribution to show the flow n to p is equal to the flow from p to n. How electron current from p to n (that is, associated with minority carries) is not affected by the height of the potential energy barrier? Explain.[10]

At thermal equilibrium (no external bias), a PN junction has a built-in potential barrier (contact potential) V₀ created by the diffusion of majority carriers. Despite this barrier, two opposing currents flow simultaneously and exactly c...

210 marksMoment of inertia and torqueAnswer

Describe moment of inertia and torque for a rotating rigid body. Find the expression for rotational kinetic energy and discuss the conditions for conservation.[10]

--- The inability of a body to change its state of rest or uniform rotational motion by itself is called inertia. For a rotating body, this property is called the Moment of Inertia. For a system of particles with masses m₁, m₂, m₃, ..., ...

310 marksBlackbody radiationAnswer

Explain the theory of black body radiation. Why this theory needs quantum mechanical interpretation? How this interpretation became experimentally successful? Explain.[10]

A black body is an ideal body that absorbs all incident radiation falling on it, regardless of wavelength or angle of incidence. When such a body is heated, it emits radiation called black body radiation (also called thermal radiation). ...

45 marksHall effectAnswer

Explain Hall effect and discuss the importance of Hall voltage while manufacturing electronic devices. [5]

The Hall Effect is the phenomenon in which a transverse electric field (and hence a potential difference) is developed across a current-carrying conductor when it is placed in a magnetic field perpendicular to the direction of current fl...

55 markseffective mass and holesAnswer

Discuss effective mass of electrons and holes. [5]

In a crystalline solid (semiconductor), electrons and holes do not move as free particles. They move through a periodic crystal lattice and experience internal forces from the lattice ions in addition to any externally applied force. To ...

65 marksElectrical conductivity of semiconductorsAnswer

Describe electrical conductivity of semiconductors. [5]

A semiconductor is a material whose electrical conductivity lies between that of a conductor and an insulator. Examples include Silicon (Si) and Germanium (Ge). --- From band theory, semiconductors have the following characteristics: - T...

75 marksNumericalOscillation of springAnswer

An oscillating block of mass 250 g takes 0.15 sec to move between the endpoints of the motion, which are 40 cm apart. (a) What is the frequency of the motion? (b) What is the amplitude of the motion? (c) What is the force constant of the spring? [5]

Solution: Oscillating Block on a Spring

STEP 1 - Given Data

Quantity	Value
Mass $m$	250 g = 0.25 kg
Time to move between endpoints	0.15 s
Distance between endpoints	40 cm = 0.40 m

Interpretation: Moving from one extreme endpoint to the other is half a full oscillation, so this time equals $T/2$ .

$\frac{T}{2} = 0.15 \text{ s} \implies T = 0.30 \text{ s}$

STEP 2 - Solve

(a) Frequency

$f = \frac{1}{T} = \frac{1}{0.30} \approx 3.33 \text{ Hz}$

(b) Amplitude

Amplitude = half the total distance between endpoints:

$A = \frac{0.40}{2} = 0.20 \text{ m} = 20 \text{ cm}$

(c) Force Constant

$T = 2\pi\sqrt{\frac{m}{k}} \implies k = \frac{4\pi^2 m}{T^2}$

$k = \frac{4 \times (3.14159)^2 \times 0.25}{(0.30)^2} = \frac{4 \times 9.8696 \times 0.25}{0.09}$

$k = \frac{9.8696}{0.09} \approx 109.7 \text{ N/m}$

Summary

Part	Result
(a) Frequency	$f \approx 3.33$ Hz
(b) Amplitude	$A = 0.20$ m
(c) Force constant	$k \approx 109.7$ N/m

All results confirmed.

Solution: Oscillating Block on a Spring

STEP 1 - Given Data

Quantity	Value
Mass $m$	250 g = 0.25 kg
Time to move between endpoints	0.15 s
Distance between endpoints	40 cm = 0.40 m

Interpretation: Moving from one extreme endpoint to the other is half a full oscillation, so this time equals $T/2$.

$$\frac{T}{2} = 0.15 \text{ s} \implies T = 0.30 \text{ s}$$

STEP 2 - Solve

(a) Frequency

$$f = \frac{1}{T} = \frac{1}{0.30} \approx 3.33 \text{ Hz}$$

(b) Amplitude

Amplitude = half the total distance between endpoints:

$$A = \frac{0.40}{2} = 0.20 \text{ m} = 20 \text{ cm}$$

(c) Force Constant

$$T = 2\pi\sqrt{\frac{m}{k}} \implies k = \frac{4\pi^2 m}{T^2}$$

$$k = \frac{4 \times (3.14159)^2 \times 0.25}{(0.30)^2} = \frac{4 \times 9.8696 \times 0.25}{0.09}$$

$$k = \frac{9.8696}{0.09} \approx 109.7 \text{ N/m}$$

Summary

Part	Result
(a) Frequency	$f \approx 3.33$ Hz
(b) Amplitude	$A = 0.20$ m
(c) Force constant	$k \approx 109.7$ N/m

All results confirmed.

85 marksNumericalHall effectAnswer

A current of 50 A is established in a slab of copper 0.5 cm thick and 2 cm wide. The slab is placed in a magnetic field B of 1.5 T. The magnetic field is perpendicular to the plane of the slab and to the current. The free electron concentration in copper is $8.4 \times 10^{28}$ electrons/m³. What will be the magnitude of the Hall voltage across the width of the slab? [5]

Hall Voltage Across a Copper Slab

STEP 1 - Given Data

Quantity	Value
Current, $I$	$50\ \text{A}$
Thickness, $t$	$0.5\ \text{cm} = 0.5\times10^{-2}\ \text{m}$
Width, $w$	$2\ \text{cm} = 2\times10^{-2}\ \text{m}$
Magnetic field, $B$	$1.5\ \text{T}$
Electron concentration, $n$	$8.4\times10^{28}\ \text{m}^{-3}$
Electron charge, $e$	$1.6\times10^{-19}\ \text{C}$

The field $B$ is perpendicular to the plane of the slab. The Hall voltage develops across the width, and the relevant thickness in the formula is the dimension parallel to $B$ , i.e. the thickness $t = 0.5$ cm.

STEP 2 - Solve

Hall voltage formula:

$V_H = \frac{BI}{net}$

Numerator:

$BI = 1.5 \times 50 = 75$

Denominator:

$net = (8.4\times10^{28})(1.6\times10^{-19})(0.5\times10^{-2})$

$= 8.4 \times 1.6 \times 0.5 \times 10^{28-19-2}$

$= 6.72 \times 10^{7}$

Therefore:

$V_H = \frac{75}{6.72\times10^{7}} = 1.116\times10^{-6}\ \text{V}$

$\boxed{V_H \approx 1.12\ \mu\text{V}}$

Result

The magnitude of the Hall voltage across the width of the copper slab is approximately $1.12\ \mu\text{V}$ .

The width (2 cm) does not enter the calculation because the Hall field is set up across the width, but the geometric factor in $V_H = BI/(net)$ uses the thickness (the dimension along $B$ ). This tiny value reflects copper's very high free-electron density.

Hall Voltage Across a Copper Slab

STEP 1 - Given Data

Quantity	Value
Current, $I$	$50\ \text{A}$
Thickness, $t$	$0.5\ \text{cm} = 0.5\times10^{-2}\ \text{m}$
Width, $w$	$2\ \text{cm} = 2\times10^{-2}\ \text{m}$
Magnetic field, $B$	$1.5\ \text{T}$
Electron concentration, $n$	$8.4\times10^{28}\ \text{m}^{-3}$
Electron charge, $e$	$1.6\times10^{-19}\ \text{C}$

The field $B$ is perpendicular to the plane of the slab. The Hall voltage develops across the width, and the relevant thickness in the formula is the dimension parallel to $B$, i.e. the thickness $t = 0.5$ cm.

STEP 2 - Solve

Hall voltage formula:

$$V_H = \frac{BI}{net}$$

Numerator:

$$BI = 1.5 \times 50 = 75$$

Denominator:

$$net = (8.4\times10^{28})(1.6\times10^{-19})(0.5\times10^{-2})$$

$$= 8.4 \times 1.6 \times 0.5 \times 10^{28-19-2}$$

$$= 6.72 \times 10^{7}$$

Therefore:

$$V_H = \frac{75}{6.72\times10^{7}} = 1.116\times10^{-6}\ \text{V}$$

$$\boxed{V_H \approx 1.12\ \mu\text{V}}$$

Result

The magnitude of the Hall voltage across the width of the copper slab is approximately $1.12\ \mu\text{V}$.

The width (2 cm) does not enter the calculation because the Hall field is set up across the width, but the geometric factor in $V_H = BI/(net)$ uses the thickness (the dimension along $B$). This tiny value reflects copper's very high free-electron density.

95 marksNumericalde Broglie's hypothesis and its experimentAnswer

The uncertainty in the position of a particle is equal to the de Broglie wavelength of particle. Calculate the uncertainty in the velocity of the particle in terms of the velocity of the de Broglie wave associated with the particle. [5]

Uncertainty in position: $\Delta x = \lambda$ (the de Broglie wavelength) - de Broglie wavelength: $\lambda = \dfrac{h}{mv}$ , where $m$ = mass, $v$ = particle velocity - Required: $\Delta v$ expressed in terms of the velocity of the de...

Uncertainty in position: $\Delta x = \lambda$ (the de Broglie wavelength) - de Broglie wavelength: $\lambda = \dfrac{h}{mv}$, where $m$ = mass, $v$ = particle velocity - Required: $\Delta v$ expressed in terms of the velocity of the de...

105 marksNumericalspace quantization and spinAnswer

(a). How many atomic states are there in hydrogen with n = 3? (b) How are they distributed among the subshells? Label each state with the appropriate set of quantum numbers n, 1, m, m (c) Show that the number of states in a shell, that is, states having the same n, is given by 2n22n^22n2. [5]

Principal quantum number: $n = 3$ - Quantum number rules: - $l = 0, 1, \ldots, (n-1)$ - $ml = -l, \ldots, 0, \ldots, +l$ → $(2l+1)$ values - $ms = \pm\tfrac{1}{2}$ → 2 values --- For $n = 3$ , $l = 0, 1, 2$ . $$N = 2n^2 = 2(3)^2 = 18 \te...

Principal quantum number: $n = 3$ - Quantum number rules: - $l = 0, 1, \ldots, (n-1)$ - $ml = -l, \ldots, 0, \ldots, +l$ → $(2l+1)$ values - $ms = \pm\tfrac{1}{2}$ → 2 values --- For $n = 3$, $l = 0, 1, 2$. $$N = 2n^2 = 2(3)^2 = 18 \te...

115 marksNumericalCrystal structureAnswer

Copper has a face-centered cubic structure with a non-atom basis. The density of copper is $8.96\text{ g/cm}^3$ and its atomic weight is $63.5\text{ g/mole}$ . What is the length of the unit cube of the structure? [5]

Copper has a face-centered cubic structure with a non-atom basis. The density of copper is $8.96\text{ g/cm}^3$ and its atomic weight is $63.5\text{ g/mole}$. What is the length of the unit cube of the structure? [5]

Quantity Value ------ Structure FCC, one-atom basis Density $\rho$ $8.96\ \text{g/cm}^3$ Atomic weight $M$ $63.5\ \text{g/mole}$ Avogadro number $NA$ $6.022\times10^{23}\ \text{/mole}$ Atoms per cell (FCC) $n$ $4$ Atoms per unit cell (FC...

Physics 2074 question paper

Solution: Oscillating Block on a Spring

STEP 1 - Given Data

STEP 2 - Solve

(a) Frequency

(b) Amplitude

(c) Force Constant

Summary

Solution: Oscillating Block on a Spring

STEP 1 - Given Data

STEP 2 - Solve

(a) Frequency

(b) Amplitude

(c) Force Constant

Summary

Hall Voltage Across a Copper Slab

STEP 1 - Given Data

STEP 2 - Solve

Result

Hall Voltage Across a Copper Slab

STEP 1 - Given Data

STEP 2 - Solve

Result

Solution: Oscillating Block on a Spring

STEP 1 - Given Data

STEP 2 - Solve

(a) Frequency

(b) Amplitude

(c) Force Constant

Summary

Hall Voltage Across a Copper Slab

STEP 1 - Given Data

STEP 2 - Solve

Result