IGNOU BPHCT-135 Solved Question Paper PDF Download

IGNOU BPHCT-135 Previous Year Solved Question Paper in Hindi

IGNOU BPHCT-135 Previous Year Solved Question Paper in English

Q1. Attempt any five parts: (a) Write Zeroth Law of Thermodynamics and state its physical significance. (b) Draw Andrew’s curves for CO2. (c) Write the expressions for $\bar{v}, v_{rms}$ and $v_p$ for a Maxwellian gas. (d) Write the expressions for enthalpy and Helmholtz energy. (e) Write the expression for $C_p – C_v$ for an ideal gas and a van der Waals’ gas. (f) State Joule-Thomson effect. Write the expression for Joule-Thomson coefficient. (g) Plot spectral energy density of a black body with wavelength at three different temperatures ($T_1 > T_2 > T_3$). (h) Write the Boltzmann relation between entropy and probability. What is its significance?

Ans. (a) Zeroth Law of Thermodynamics: The law states that “If two systems, A and B, are separately in thermal equilibrium with a third system, C, then systems A and B are also in thermal equilibrium with each other.” Physical Significance: The primary significance of this law is that it introduces the concept of temperature as a fundamental and measurable property of matter. It implies that all bodies in thermal equilibrium share a common property, which we call temperature. This principle is the basis for the functioning of thermometers.

(b) Andrew’s Curves for CO2: Andrew’s curves for CO2 are isotherms plotted on a Pressure-Volume (P-V) diagram at different constant temperatures. In the curves:

• $T_c$ is the critical temperature . Above this temperature, the gas cannot be liquefied by increasing pressure alone.
• Below $T_c$, the horizontal part of the curve represents the co-existence of the gas and liquid phases.
• Point P is the critical point , where the distinction between the liquid and gas phases vanishes.

(c) Velocities for a Maxwellian Gas: For a Maxwellian gas (an ideal gas), the expressions for the speeds of its molecules are as follows, where k is the Boltzmann constant, T is the absolute temperature, and m is the mass of one molecule.

• Most probable speed ($v_p$): $v_p = \sqrt{\frac{2kT}{m}}$
• Average speed ($\bar{v}$): $\bar{v} = \sqrt{\frac{8kT}{\pi m}}$
• Root-mean-square speed ($v_{rms}$): $v_{rms} = \sqrt{\frac{3kT}{m}}$

(d) Enthalpy and Helmholtz Energy: These are thermodynamic potentials with the following expressions:

• Enthalpy (H): Also known as heat content, it is a measure of the total energy of a system. Its expression is: $H = U + PV$ where U is the internal energy, P is the pressure, and V is the volume.
• Helmholtz Free Energy (F): It is a measure of the useful work obtainable from a closed system at constant temperature and volume. Its expression is: $F = U – TS$ where U is the internal energy, T is the temperature, and S is the entropy.

(e) Expression for $C_p – C_v$:

• For an Ideal Gas: According to Mayer’s relation, the difference between the molar specific heat at constant pressure ($C_p$) and molar specific heat at constant volume ($C_v$) for an ideal gas is equal to the universal gas constant (R). $C_p – C_v = R$
• For a van der Waals’ Gas: For a real gas that obeys the van der Waals’ equation, this difference is not just R but also depends on the intermolecular forces of attraction. $C_p – C_v \approx R \left(1 + \frac{2a}{RTV}\right)$ where ‘a’ is the van der Waals constant representing intermolecular attraction, T is the temperature, and V is the molar volume.

(f) Joule-Thomson Effect: Statement: When a gas is allowed to expand adiabatically from a region of high pressure to a region of low pressure through a porous plug or a small aperture, a change in its temperature is observed. This phenomenon is called the Joule-Thomson effect . Joule-Thomson Coefficient ($\mu_{JT}$): It is the rate of change of temperature with pressure at constant enthalpy. $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$ If $\mu_{JT}$ is positive, cooling occurs upon expansion; if it is negative, heating occurs.

(g) Black Body Radiation Plot: The plot of spectral energy density ($\rho_\lambda$) versus wavelength ($\lambda$) for a black body at three different temperatures ($T_1 > T_2 > T_3$) is shown below. Key observations from the plot:

• As temperature increases, the energy emitted at every wavelength increases.
• The wavelength ($\lambda_m$) at which the maximum energy is emitted shifts to shorter wavelengths as temperature increases (Wien’s Displacement Law).
• The total area under the curve (total energy radiated) is proportional to the fourth power of the temperature (Stefan-Boltzmann Law).

(h) Boltzmann Relation between Entropy and Probability: The Boltzmann relation between entropy (S) and thermodynamic probability (W) is given by: $S = k_B \ln W$ where S is the entropy of the system, $k_B$ is the Boltzmann constant, and W is the number of microstates corresponding to that macrostate. Significance: This equation forms a fundamental bridge between thermodynamics (a macroscopic theory) and statistical mechanics (a microscopic theory). It provides a physical interpretation of entropy: entropy is a measure of the disorder or randomness of a system. The Second Law of Thermodynamics, which states that the entropy of an isolated system always increases, implies that a system naturally evolves towards a more probable or more disordered state.

Q2. Answer any two parts: (a) Obtain an expression for survival equation for distribution of free paths. (b) Define coefficient of viscosity. Discuss the effect of temperature and pressure on viscosity. (c) Show that sedimentation is an example of Brownian motion. Discuss its utility in daily life.

Ans. (a) Survival Equation for Distribution of Free Paths: In a gas, a molecule moves in a straight line between two successive collisions. This distance is called a free path. The survival equation gives the probability that a molecule will travel a distance x without suffering a collision. Let $\lambda$ be the mean free path. This means the probability of a molecule having a collision while traveling a unit distance is $1/\lambda$. Let $P(x)$ be the probability that a molecule travels a distance x without a collision. The probability that this molecule will collide in the next small distance dx is $(dx/\lambda)$. Therefore, the probability of surviving a distance x+dx without a collision is the product of the probability of surviving distance x and the probability of not colliding in dx . $P(x+dx) = P(x) \times (1 – \frac{dx}{\lambda})$ $P(x+dx) – P(x) = -P(x) \frac{dx}{\lambda}$ $dP(x) = -P(x) \frac{dx}{\lambda}$ $\frac{dP(x)}{P(x)} = -\frac{dx}{\lambda}$ Integrating this equation: $\int_{P(0)}^{P(x)} \frac{dP(x)}{P(x)} = -\int_0^x \frac{dx}{\lambda}$ The probability of traveling a distance $x=0$ without collision is 1, so $P(0)=1$. $[\ln P(x)]_1^{P(x)} = -[x/\lambda]_0^x$ $\ln P(x) – \ln 1 = -x/\lambda$ $\ln P(x) = -x/\lambda$ $P(x) = e^{-x/\lambda}$ This is the survival equation . If we start with $N_0$ molecules, the number of molecules $N(x)$ that have not had a collision after traveling a distance x is: $N(x) = N_0 e^{-x/\lambda}$ This equation shows that the distribution of free paths is exponential.

(b) Coefficient of Viscosity and Effect of Temperature and Pressure: Definition: The coefficient of viscosity ($\eta$) is a measure of a fluid’s (liquid or gas) internal resistance to flow. According to Newton’s law of viscosity, the tangential stress (force per unit area) between two parallel layers of a fluid is proportional to the velocity gradient ($dv/dz$). $F/A = -\eta \frac{dv}{dz}$ Here, the coefficient of viscosity ($\eta$) is defined as the tangential force per unit area required to maintain a unit velocity gradient in the fluid. Its SI unit is the Pascal-second (Pa.s). Effect of Temperature:

• In Gases: The viscosity of gases increases with increasing temperature. Viscosity in gases arises from the transfer of momentum. At higher temperatures, molecules move faster, leading to more effective momentum transfer and thus higher viscosity. According to kinetic theory, $\eta \propto \sqrt{T}$.
• In Liquids: The viscosity of liquids decreases sharply with increasing temperature. Viscosity in liquids is primarily due to intermolecular cohesive forces. As temperature increases, the kinetic energy of molecules increases, allowing them to overcome these forces and slide past each other more easily, thus reducing viscosity.

Effect of Pressure:

• In Gases: At moderate pressures, the viscosity of a gas is almost independent of pressure. This is because an increase in pressure increases the density (number of collisions), but it decreases the mean free path, and these two effects nearly cancel each other out. A dependence is observed at very high or very low pressures.
• In Liquids: The viscosity of liquids increases slightly with an increase in pressure. High pressure forces molecules closer together, strengthening intermolecular forces and increasing the resistance to flow.

(c) Sedimentation, Brownian Motion, and its Utility: Sedimentation as an example of Brownian motion: Brownian motion is the random, erratic movement of microscopic particles suspended in a fluid. This motion is caused by the continuous and unbalanced bombardment of the particles by the molecules of the fluid. Sedimentation is the process of particles settling down under the influence of gravity. When microscopic particles are suspended in a fluid, two main forces act on them:

1. A downward gravitational force, which tends to make them settle.
2. An effective upward force due to the random collisions from the fluid molecules (Brownian motion), which tends to keep them suspended.

An equilibrium is established between these two opposing effects. As a result, the particles do not settle completely but form a vertical concentration gradient, with the density of particles decreasing with height. This equilibrium is a direct demonstration of the existence of Brownian motion, which counteracts the systematic effect of gravity. Jean Perrin’s successful determination of Avogadro’s number by studying this equilibrium proved that sedimentation is indeed a macroscopic manifestation of Brownian motion.

Utility in daily life:

• Medicine and Biochemistry: Sedimentation principles are used to separate macromolecules of different molecular weights (like proteins, DNA) using an ultracentrifuge.
• Environmental Science: It is crucial for understanding the distribution and deposition of pollutants (like dust, soot, silt) in the air and water.
• Food Industry: The separation of cream from milk or the settling of pulp in fruit juices are examples of sedimentation.
• Colloidal Chemistry: It helps in understanding the stability of paints, inks, and other colloidal solutions, where Brownian motion prevents the particles from settling out.

Q3. Answer any two parts: (a) Derive equation of state for a thermodynamic system assuming volume as a function of pressure and temperature. Imposing the condition dV=0 and dT≠0, prove that the thermodynamic variables are connected through a relation : $(\frac{\partial P}{\partial T})_V (\frac{\partial T}{\partial V})_P (\frac{\partial V}{\partial P})_T = -1$ (b) 1 g of water and steam at normal atmospheric pressure occupy 1 cm³ and 1671 cm³ volume, respectively. Calculate the change in internal energy for vaporisation of water at 100°C. Take latent heat of steam as 540 cal/g and J = 4.2 J/cal. (c) Obtain Laplace’s formula for speed of sound.

Ans. (a) Derivation of the Relation between Thermodynamic Variables: For a thermodynamic system, the state variables Pressure (P), Volume (V), and Temperature (T) are connected by an equation of state, which can be generally written as $f(P, V, T) = 0$. This means we can express any one variable as a function of the other two. Let us assume volume is a function of pressure and temperature: $V = V(P, T)$. The total differential is: $dV = \left(\frac{\partial V}{\partial P}\right)_T dP + \left(\frac{\partial V}{\partial T}\right)_P dT$ … (1) Similarly, we can consider pressure as a function of volume and temperature: $P = P(V, T)$. The total differential is: $dP = \left(\frac{\partial P}{\partial V}\right)_T dV + \left(\frac{\partial P}{\partial T}\right)_V dT$ … (2) Now, substitute the expression for $dP$ from equation (2) into equation (1): $dV = \left(\frac{\partial V}{\partial P}\right)_T \left[ \left(\frac{\partial P}{\partial V}\right)_T dV + \left(\frac{\partial P}{\partial T}\right)_V dT \right] + \left(\frac{\partial V}{\partial T}\right)_P dT$ Rearranging the terms: $dV = \left(\frac{\partial V}{\partial P}\right)_T \left(\frac{\partial P}{\partial V}\right)_T dV + \left(\frac{\partial V}{\partial P}\right)_T \left(\frac{\partial P}{\partial T}\right)_V dT + \left(\frac{\partial V}{\partial T}\right)_P dT$ Since $\left(\frac{\partial V}{\partial P}\right)_T = 1 / \left(\frac{\partial P}{\partial V}\right)_T$, the first term becomes $dV$. $dV = dV + \left[ \left(\frac{\partial V}{\partial P}\right)_T \left(\frac{\partial P}{\partial T}\right)_V + \left(\frac{\partial V}{\partial T}\right)_P \right] dT$ $0 = \left[ \left(\frac{\partial V}{\partial P}\right)_T \left(\frac{\partial P}{\partial T}\right)_V + \left(\frac{\partial V}{\partial T}\right)_P \right] dT$ For the given condition $dT \neq 0$, the term inside the bracket must be zero: $\left(\frac{\partial V}{\partial P}\right)_T \left(\frac{\partial P}{\partial T}\right)_V + \left(\frac{\partial V}{\partial T}\right)_P = 0$ $\left(\frac{\partial V}{\partial P}\right)_T \left(\frac{\partial P}{\partial T}\right)_V = – \left(\frac{\partial V}{\partial T}\right)_P$ Rearranging the terms, and noting that $(\frac{\partial V}{\partial T})_P = 1 / (\frac{\partial T}{\partial V})_P$ : $\left(\frac{\partial P}{\partial T}\right)_V \frac{1}{(\frac{\partial T}{\partial V})_P} \left(\frac{\partial V}{\partial P}\right)_T = -1$ Thus, we get the cyclic relation: $\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial T}{\partial V}\right)_P \left(\frac{\partial V}{\partial P}\right)_T = -1$ The condition $dV=0$ indicates we are considering an isochoric process, for which $(\partial P / \partial T)_V$ is an important physical quantity.

(b) Calculation of Change in Internal Energy: Given:

• Mass of water, $m = 1$ g
• Initial volume (of water), $V_1 = 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3$
• Final volume (of steam), $V_2 = 1671 \text{ cm}^3 = 1671 \times 10^{-6} \text{ m}^3$
• Pressure, $P =$ normal atmospheric pressure $= 1.013 \times 10^5 \text{ N/m}^2$ (Pa)
• Latent heat of steam, $L = 540 \text{ cal/g}$
• Mechanical equivalent of heat, $J = 4.2 \text{ J/cal}$

1. Calculate the heat absorbed (Q):

The heat required to convert 1g of water at 100°C to steam:

$Q = m \times L = 1 \text{ g} \times 540 \text{ cal/g} = 540 \text{ cal}$

Converting this to Joules:

$Q = 540 \text{ cal} \times 4.2 \text{ J/cal} = 2268 \text{ J}$

2. Calculate the work done (W):

Vaporization is a constant pressure process. Thus, the work done by the system is:

$W = P \times (V_2 – V_1)$

$W = (1.013 \times 10^5 \text{ N/m}^2) \times (1671 \times 10^{-6} – 1 \times 10^{-6}) \text{ m}^3$

$W = (1.013 \times 10^5) \times (1670 \times 10^{-6}) \text{ J}$

$W = 1.013 \times 1.670 \times 10^2 \text{ J} = 169.171 \text{ J}$

3. Calculate the change in internal energy ($\Delta U$):

According to the First Law of Thermodynamics:

$\Delta U = Q – W$

$\Delta U = 2268 \text{ J} – 169.171 \text{ J}$

$\Delta U = 2098.829 \text{ J}$

Therefore, the change in internal energy for the vaporisation of water is

2098.83 J

.

(c) Laplace’s Formula for Speed of Sound: Sound waves propagate through a medium as compressions and rarefactions. The speed of sound depends on the modulus of elasticity (B) and the density ($\rho$) of the medium: $v = \sqrt{B/\rho}$. Newton’s Assumption: Newton assumed that the propagation of sound is a slow, isothermal process , because the heat generated during compression dissipates into the surroundings and the cooling during rarefaction is compensated by heat from the surroundings. In this case, the modulus of elasticity is the isothermal bulk modulus ($B_T$). For an ideal gas, $PV = \text{constant}$ in an isothermal process. Differentiating: $P dV + V dP = 0 \implies P = -V \frac{dP}{dV}$ Since $B_T = -V (\frac{\partial P}{\partial V})_T$, we get $B_T = P$. Hence, Newton’s formula: $v = \sqrt{\frac{P}{\rho}}$. This formula gives a value about 16% lower than experimental values. Laplace’s Correction: Laplace argued that the propagation of sound waves is very fast. The compressions and rarefactions occur so rapidly that there is not enough time for heat exchange to take place. Therefore, the process is adiabatic , not isothermal. In this case, the modulus of elasticity is the adiabatic bulk modulus ($B_S$). For an ideal gas, $PV^\gamma = \text{constant}$ in an adiabatic process, where $\gamma = C_p/C_v$. Differentiating: $P(\gamma V^{\gamma-1} dV) + V^\gamma dP = 0$ Dividing by $V^{\gamma-1}$: $P\gamma dV + V dP = 0 \implies \gamma P = -V \frac{dP}{dV}$ Since $B_S = -V (\frac{\partial P}{\partial V})_S$, we get $B_S = \gamma P$. Substituting this value into the speed formula gives Laplace’s formula : $v = \sqrt{\frac{B_S}{\rho}} = \sqrt{\frac{\gamma P}{\rho}}$ This corrected formula agrees very well with experimental values.

Q4. When two phases of a substance coexist in equilibrium at constant temperature and pressure, their specific Gibbs’ free energies are equal. Using this fact, obtain Clausius-Clapeyron equation. (OR) Draw the indicator diagram of a Carnot cycle. Obtain the expression for the work done during one cycle of Carnot engine and hence deduce its efficiency.

Ans. Derivation of the Clausius-Clapeyron Equation When two phases of a substance (e.g., liquid and vapor) are in thermal equilibrium at a constant temperature T and pressure P, their specific Gibbs free energies (Gibbs free energy per unit mass) are equal. Let $g_1$ and $g_2$ be the specific Gibbs free energies for phase 1 (e.g., liquid) and phase 2 (e.g., vapor) respectively. At equilibrium: $g_1 = g_2$ … (1) Now, if we slightly change the temperature from T to T+dT and pressure from P to P+dP, such that the system remains in equilibrium, the new Gibbs free energies will also be equal: $g_1 + dg_1 = g_2 + dg_2$ Using equation (1), this implies: $dg_1 = dg_2$ … (2) From the fundamental thermodynamic relation, the expression for a change in Gibbs free energy is: $dG = V dP – S dT$ For specific quantities (per unit mass), this becomes: $dg = v dP – s dT$ where $v$ is the specific volume and $s$ is the specific entropy. Using this expression for $dg_1$ and $dg_2$ in equation (2): $v_1 dP – s_1 dT = v_2 dP – s_2 dT$ Rearranging the terms: $(s_2 – s_1) dT = (v_2 – v_1) dP$ This can be rearranged to give the slope of the pressure-temperature curve: $\frac{dP}{dT} = \frac{s_2 – s_1}{v_2 – v_1}$ … (3) During a phase transition, which is a reversible, isothermal process, the change in entropy $\Delta s = s_2 – s_1$ can be expressed in terms of the latent heat (L) and temperature (T): $\Delta s = \frac{L}{T}$ where L is the specific latent heat. Substituting this value into equation (3): $\frac{dP}{dT} = \frac{L}{T(v_2 – v_1)}$ This is the Clausius-Clapeyron equation . It relates the slope of the phase transition curve (e.g., boiling point curve) to the thermodynamic properties of the substance (latent heat, change in volume).

OR

Carnot Cycle

The Carnot cycle is an ideal, reversible thermodynamic cycle that provides the maximum possible efficiency for a heat engine. It consists of four processes:

1. Indicator Diagram:

The P-V indicator diagram for a Carnot cycle is shown below:

• Process 1 → 2 (Isothermal Expansion): The working substance (an ideal gas) is in contact with a heat source at $T_1$ and absorbs heat $Q_1$, causing it to expand from volume $V_1$ to $V_2$.
• Process 2 → 3 (Adiabatic Expansion): The gas is insulated and expands adiabatically, causing its temperature to drop from $T_1$ to $T_2$ and its volume to increase from $V_2$ to $V_3$.
• Process 3 → 4 (Isothermal Compression): The gas is brought into contact with a heat sink at $T_2$ and is compressed, causing it to reject heat $Q_2$ to the sink. The volume decreases from $V_3$ to $V_4$.
• Process 4 → 1 (Adiabatic Compression): The gas is insulated again and compressed adiabatically, raising its temperature back to $T_1$ and returning it to its initial state.

2. Expression for Work Done:

The net work done in one cycle is the sum of the work done in the four processes.

$W_{1\to2}$ (Isothermal Expansion) = $nRT_1 \ln\left(\frac{V_2}{V_1}\right) = Q_1$

$W_{2\to3}$ (Adiabatic Expansion) = $\frac{nR}{\gamma-1}(T_1 – T_2)$

$W_{3\to4}$ (Isothermal Compression) = $nRT_2 \ln\left(\frac{V_4}{V_3}\right) = -nRT_2 \ln\left(\frac{V_3}{V_4}\right)$

$W_{4\to1}$ (Adiabatic Compression) = $\frac{nR}{\gamma-1}(T_2 – T_1) = -W_{2\to3}$

The net work $W_{net} = W_{1\to2} + W_{2\to3} + W_{3\to4} + W_{4\to1}$. The adiabatic work terms cancel each other out.

$W_{net} = nRT_1 \ln\left(\frac{V_2}{V_1}\right) + nRT_2 \ln\left(\frac{V_4}{V_3}\right)$

For the adiabatic processes (2→3 and 4→1), we have:

$T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1}$ and $T_1 V_1^{\gamma-1} = T_2 V_4^{\gamma-1}$

Dividing these equations gives: $\left(\frac{V_2}{V_1}\right)^{\gamma-1} = \left(\frac{V_3}{V_4}\right)^{\gamma-1} \implies \frac{V_2}{V_1} = \frac{V_3}{V_4}$

Using this relation, we can simplify the expression for work:

$W_{net} = nRT_1 \ln\left(\frac{V_2}{V_1}\right) – nRT_2 \ln\left(\frac{V_2}{V_1}\right)$

$W_{net} = nR(T_1 – T_2) \ln\left(\frac{V_2}{V_1}\right)$

3. Efficiency ($\eta$):

The efficiency of an engine is defined as the ratio of the net work done to the heat absorbed from the source.

$\eta = \frac{\text{Net Work Done}}{\text{Heat Input}} = \frac{W_{net}}{Q_1}$

We know that $Q_1 = nRT_1 \ln\left(\frac{V_2}{V_1}\right)$.

$\eta = \frac{nR(T_1 – T_2) \ln\left(\frac{V_2}{V_1}\right)}{nRT_1 \ln\left(\frac{V_2}{V_1}\right)}$

$\eta = \frac{T_1 – T_2}{T_1} = 1 – \frac{T_2}{T_1}$

This is the expression for the efficiency of a Carnot engine, which depends only on the absolute temperatures of the source and the sink.

Q5. Starting from the expression for thermodynamic probability given by: $W[N_i] = \prod_i \frac{(N_i+g_i-1)!}{N_i!(g_i-1)!}$ obtain the expression for Bose-Einstein distribution function. (OR) Define Fermi energy. Obtain its expression. Calculate the value of Fermi energy for copper. (Given that: $n = 8.53 \times 10^{28}$ electrons per $m^3$, $m_e = 9.1 \times 10^{-31}$ kg and $h = 6.62 \times 10^{-34}$ Js)

Ans. Derivation of Bose-Einstein Distribution Function Bose-Einstein statistics applies to particles (bosons) that are indistinguishable and do not obey the Pauli exclusion principle, meaning more than one particle can occupy the same quantum state. The number of ways (thermodynamic probability) to distribute $N_i$ particles into the $i$-th energy level, which has energy $\epsilon_i$ and degeneracy $g_i$, is given by: $W = \prod_i \frac{(N_i+g_i-1)!}{N_i!(g_i-1)!}$ We want to find the most probable distribution, which maximizes W. It is mathematically easier to maximize $\ln W$ instead of W. $\ln W = \sum_i [\ln(N_i+g_i-1)! – \ln N_i! – \ln(g_i-1)!]$ Using Stirling’s approximation ($\ln x! \approx x \ln x – x$) and assuming $N_i$ and $g_i \gg 1$, so that $g_i-1 \approx g_i$ and $N_i+g_i-1 \approx N_i+g_i$. $\ln W \approx \sum_i [ (N_i+g_i)\ln(N_i+g_i) – (N_i+g_i) – (N_i\ln N_i – N_i) – (g_i\ln g_i – g_i) ]$ $\ln W \approx \sum_i [ (N_i+g_i)\ln(N_i+g_i) – N_i\ln N_i – g_i\ln g_i ]$ For the most probable distribution, $\delta(\ln W)=0$, subject to the two constraints: 1. Total number of particles is constant: $N = \sum N_i \implies \delta N = \sum \delta N_i = 0$ 2. Total energy is constant: $U = \sum N_i \epsilon_i \implies \delta U = \sum \epsilon_i \delta N_i = 0$ Using Lagrange’s method of undetermined multipliers: $\delta(\ln W) – \alpha \sum \delta N_i – \beta \sum \epsilon_i \delta N_i = 0$ $\sum_i \left[ \frac{\partial(\ln W)}{\partial N_i} – \alpha – \beta\epsilon_i \right] \delta N_i = 0$ Differentiating $\ln W$ with respect to $N_i$: $\frac{\partial(\ln W)}{\partial N_i} = [\ln(N_i+g_i) + \frac{N_i+g_i}{N_i+g_i}] – [\ln N_i + \frac{N_i}{N_i}] = \ln(N_i+g_i) – \ln N_i = \ln\left(\frac{N_i+g_i}{N_i}\right)$ Substituting this into the Lagrange equation: $\ln\left(\frac{N_i+g_i}{N_i}\right) – \alpha – \beta\epsilon_i = 0$ $\ln\left(1 + \frac{g_i}{N_i}\right) = \alpha + \beta\epsilon_i$ $1 + \frac{g_i}{N_i} = e^{\alpha + \beta\epsilon_i}$ $\frac{g_i}{N_i} = e^{\alpha + \beta\epsilon_i} – 1$ The Bose-Einstein distribution function, $f_{BE}(\epsilon_i)$, is defined as the average number of particles per quantum state in the energy level $i$, i.e., $f_{BE}(\epsilon_i) = N_i/g_i$. $f_{BE}(\epsilon_i) = \frac{N_i}{g_i} = \frac{1}{e^{\alpha + \beta\epsilon_i} – 1}$ where $\beta = 1/k_B T$ and $\alpha = -\mu/k_B T$ ($\mu$ is the chemical potential). Thus, the Bose-Einstein distribution function is: $f_{BE}(\epsilon) = \frac{1}{e^{(\epsilon – \mu)/k_B T} – 1}$

OR

Fermi Energy and its Calculation

Definition:

The Fermi energy ($E_F$) is defined as the energy of the highest occupied quantum state in a system of fermions (like electrons in a metal) at absolute zero temperature (0 K). At absolute zero, all energy states below the Fermi energy are filled, and all states above it are empty.

Derivation of Expression:

Consider a system of N free electrons in a volume V. The number of quantum states in the momentum interval $p$ to $p+dp$, accounting for spin (2 states), is given by:

$g(p)dp = \frac{2V}{h^3} (4\pi p^2 dp)$

At absolute zero (T=0 K), electrons fill the lowest available energy levels according to the Pauli exclusion principle. All levels are filled up to a maximum momentum, the Fermi momentum $p_F$.

The total number of electrons N is therefore:

$N = \int_0^{p_F} g(p) dp = \int_0^{p_F} \frac{8\pi V}{h^3} p^2 dp$

$N = \frac{8\pi V}{h^3} \left[ \frac{p^3}{3} \right]_0^{p_F} = \frac{8\pi V}{3h^3} p_F^3$

Rearranging for $p_F$:

$p_F^3 = \frac{3N h^3}{8\pi V} \implies p_F = h \left( \frac{3n}{8\pi} \right)^{1/3}$

where $n=N/V$ is the electron density.

The Fermi energy $E_F$ is the kinetic energy corresponding to the Fermi momentum $p_F$:

$E_F = \frac{p_F^2}{2m_e} = \frac{1}{2m_e} \left[ h \left( \frac{3n}{8\pi} \right)^{1/3} \right]^2$

$E_F = \frac{h^2}{2m_e} \left( \frac{3n}{8\pi} \right)^{2/3}$

This is the expression for the Fermi energy.

Calculation for Copper:

Given:

• Electron density, $n = 8.53 \times 10^{28} \text{ m}^{-3}$ (Note: The value in the English paper text `10^7*` is a typo; the correct value from the Hindi text is used).
• Mass of electron, $m_e = 9.1 \times 10^{-31} \text{ kg}$
• Planck’s constant, $h = 6.62 \times 10^{-34} \text{ Js}$

Substituting the values into the expression for Fermi energy:

$E_F = \frac{(6.62 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31}} \left( \frac{3 \times 8.53 \times 10^{28}}{8\pi} \right)^{2/3}$

$E_F = \frac{43.8244 \times 10^{-68}}{18.2 \times 10^{-31}} \left( \frac{25.59 \times 10^{28}}{25.133} \right)^{2/3}$

$E_F = (2.4079 \times 10^{-37}) \times (1.0182 \times 10^{28})^{2/3}$

$E_F = (2.4079 \times 10^{-37}) \times (4.70 \times 10^{18})$

$E_F = 11.32 \times 10^{-19} \text{ J}$

To convert the energy to electron-volts (eV), we divide by the elementary charge $1.602 \times 10^{-19} \text{ J/eV}$:

$E_F (\text{eV}) = \frac{11.32 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 7.07 \text{ eV}$

Thus, the value of Fermi energy for copper is approximately

7.07 eV

.