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Consider an isolated system of ideal gas of $N$ molecules
contained in a volume $V$ and having an energy
$$E=\sum_{i=1}^{3N}\frac{p_i^2}{2m}.$$ Show that the number of states in the
energy range $U-\Delta$ and $U$ is
of the system is given by\hfill [5]
\begin{eqnarray*}
\frac{1}{\Gamma(3N/2+1)}\frac{3N\Delta}{2U} \Big(\frac{mU V^{2/3}
}{2\pi\hbar^2}\Big)^{3N/2}
\end{eqnarray*}

Quantum effects in macroscopic systems appear in two ways. The first the energy levels are quantized. The quantization of energy levels does not need any modification in the framework. Secondly identical nature of particles constituting the system. This requires a new approach to enumerating the microstates. The microstates are not labeled by coordinates and momenta as is the case in classical theory. In quantum theory the microstates are specified by giving the number of particles for different levels.

$\newcommand{\DD}[2][]{\frac{d^2 #1}{d^2 #2}}
\newcommand{\matrixelement}[3]{\langle#1|#2|#3\rangle}\newcommand{\PP}[2][]{\frac{\partial^2 #1}{\partial #2^2}}\newcommand{\dd}[2][]{\frac{d#1}{d#2}}\newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}\newcommand{\average}[2]{\langle#1|#2|#1\rangle} \newcommand{\dd}[2][]{\frac{d#1}{d#2}}\newcommand{\Label}[1]{\label{#1}}$

We discuss the solution of time dependent one particle Schrodinger equation and obtain an expression for the propagator giving the time development.

\begin{eqnarray} \frac{\mu V_0}{\hbar^2 k}\left|\int_0^{R_0} \left( e^{2ikr}-1\right) dr\right|  &=& \frac{\mu V_0}{\hbar^2 k}\left|\frac{e^{2ikR_0}-1}{2ik} - R_0 \right|\label{E2}\\ &=& \frac{\mu V_0}{2\hbar^2 k^2}\left|e^{2ikR_0}- 2ik R_0 -1 \right|\label{E3} \end{eqnarray}

A cylinder closed at both ends with adiabatic walls, is divided into two parts by a movable piston. The piston is frictionless and adiabatic. Originally, the pressure, volume, and the temperature of the gas are the same, $(P_0,V_0,T_0)$, on the two sides of the piston. The gas is ideal gas with $C_v$ independent of $T$ and $\gamma=1.5$. By means of a heating coil on the left hand side, heat is slowly supplied to the gas on the left hand side until the pressure reaches ${27\over 8}P_0$.

1. what is the entropy change of the gas on the left?
2. what is the entropy change of the gas on the right?

Electrodynamics                                              Apr 19, 2019
                                     Quiz-IX

The above image is reproduced from a book. Read Example 8.5 carefully. Answer the following questions.

1. Do you agree that there is no flux linked with the circuit when key is open?
2. Do you agree that there will be induced current when the circuit is closed. But the coil and the magnet remain stationary?
3. If you disagree what is the mistake? Does the flux rule apply or not in this case?
4. Write any other comment you may have.

Thomson passed electrons through a region having mutually perpendicular electric and magnetic field, and both perpendicular to the velocity of the electrons. The fields were adjusted so as to produce no deflection. This enebled him to measure the \(e/m\) of electrons.