Infinite Square Well Energy Levels
The energy eigenvalues for a particle in a 2D infinite square well of side length $L$ are given by the formula:
$ E_{n_x, n_y} = \frac{\pi^2 \hbar^2}{2mL^2} (n_x^2 + n_y^2) $
where $n_x$ and $n_y$ are positive integers ($n_x, n_y = 1, 2, 3, ...$) representing the quantum states along the x and y directions, respectively.
Determining Quantum Numbers
The wavefunction is given by:
$ \psi_{n_x, n_y}(x, y) = \left(\frac{2}{L}\right) \sin\left(\frac{n_x \pi x}{L}\right) \sin\left(\frac{n_y \pi y}{L}\right) $
We are told the wavefunction is zero along the line $y = \frac{L}{2}$ (apart from the boundaries $y=0$ and $y=L$). This condition requires:
$ \sin\left(\frac{n_y \pi (L/2)}{L}\right) = 0 $
$ \sin\left(\frac{n_y \pi}{2}\right) = 0 $
This equation holds true when $\frac{n_y \pi}{2}$ is an integer multiple of $\pi$. Thus, $\frac{n_y}{2} = k$, where $k$ is an integer. This means $n_y = 2k$. Since $n_y$ must be a positive integer, the smallest possible value for $n_y$ is 2 (when $k=1$).
The problem states the wavefunction is zero *only* along $y = L/2$ (besides boundaries), implying the simplest state satisfying this condition. The simplest choice for $n_x$ is $n_x = 1$. Therefore, the quantum numbers for the given state are $(n_x, n_y) = (1, 2)$.
Calculating Energies
The energy $E$ corresponds to the state $(1, 2)$:
$ E = E_{1,2} = \frac{\pi^2 \hbar^2}{2mL^2} (1^2 + 2^2) = \frac{\pi^2 \hbar^2}{2mL^2} (1 + 4) = 5 \frac{\pi^2 \hbar^2}{2mL^2} $
The ground state energy corresponds to the quantum numbers $(n_x, n_y) = (1, 1)$:
$ E_{ground} = E_{1,1} = \frac{\pi^2 \hbar^2}{2mL^2} (1^2 + 1^2) = 2 \frac{\pi^2 \hbar^2}{2mL^2} $
Relating Energies
We can express the ground state energy in terms of $E$. From the expression for $E$, we have:
$ \frac{\pi^2 \hbar^2}{2mL^2} = \frac{E}{5} $
Substituting this into the expression for the ground state energy:
$ E_{ground} = 2 \left( \frac{\pi^2 \hbar^2}{2mL^2} \right) = 2 \left( \frac{E}{5} \right) = \frac{2}{5}E $
Thus, the energy of the ground state is $\frac{2}{5}E$. This corresponds to Option B.
Important Questions from Schrödinger Equation 1D Potentials Harmonic Oscillator
The wavefunction of a particle in an infinite one-dimensional potential well at time $t$ is
$\Psi(x, t) = \sqrt{\frac{2}{3}} e^{-iE_1t/\hbar}\psi_1(x) + \frac{1}{\sqrt{6}} e^{i\pi/6}e^{-iE_2t/\hbar}\psi_2(x) + \frac{1}{\sqrt{6}} e^{i\pi/4}e^{-iE_3t/\hbar}\psi_3(x)$
where $\psi_1, \psi_2$ and $\psi_3$ are the normalized ground state, the normalized first excited state and the normalized second excited state, respectively. $E_1, E_2$ and $E_3$ are the eigen-energies corresponding to $\psi_1, \psi_2$ and $\psi_3$, respectively. The expectation value of energy of the particle in state $\Psi(x, t)$ isA particle is subjected to a potential
$V(x) = \begin{cases} \infty, & x \le 0 \\ V_0, & a \le x \le b \\ 0, & \text{elsewhere} \end{cases}$
Here, $a > 0$ and $b > a$. If the energy of the particle $E < V_0$, which one of the following schematics is a valid quantum mechanical wavefunction ($\Psi$) for the system?- The wavefunction for a particle is given by the form $e^{-(iax+\beta)}$, where $a$ and $\beta$ are real constants. In which one of the following potentials $V(x)$, the particle is moving?
A particle of mass $m$ is moving in the potential
$V(x) = \begin{cases} V_0 + \frac{1}{2}m\omega_0^2x^2, & x > 0, \\ \infty, & x \le 0, \end{cases}$
Figures P, Q, R and S show different combinations of the values of $\omega_0$ and $V_0$.
$E_j^{(P)}, E_j^{(Q)}, E_j^{(R)}$ and $E_j^{(S)}$ with $j = 0, 1, 2, ...$, are the eigen-energies of the $j$-th level for the potentials shown in figures P, Q, R and S, respectively. Which of the statement is/are true?- Young's double slit experiment is performed using a beam of $C_{60}$ (fullerene) molecules, each molecule being made up of 60 carbon atoms. When the slit separation is 50 nm, fringes are formed on a screen kept at a distance of 1 m from the slits. Now, the experiment is repeated with $C_{70}$ molecules with a slit separation of 92.5 nm. The kinetic energies of both the beams are the same. The position of the 4th bright fringe for $C_{60}$ will correspond to the $n^{th}$ bright fringe for $C_{70}$. What is the value of $n$ (rounded off to the nearest integer) ?
