$V(x,y)=0$ for $-a<x<a$ and $-a<y<a$
$= \infty$ elsewhere
The energy of the first excited state for this particle is given by,
2D Square Well Particle Energy Basics
The problem asks for the energy of the first excited state of a particle with mass $m$ confined to a two-dimensional square well potential. The energy levels in such a system are determined by the well's dimensions and quantum numbers.
Energy Formula Calculation
The energy eigenvalues for a particle in a 2D square well of side length $L$ are given by:
$ E_{n_x, n_y} = \frac{\pi^2\hbar^2}{2mL^2} (n_x^2 + n_y^2) $Where $\hbar$ is the reduced Planck constant, and $n_x, n_y$ are positive integers ($n_x \ge 1, n_y \ge 1$) representing the quantum states. To match the provided options, we interpret the well dimension 'a' in the question as the side length, so $L=a$.
First Excited State Energy Derivation
We calculate the energies for the lowest quantum states:
- Ground State ($n_x=1, n_y=1$): This is the state with the minimum energy. $ E_{1,1} = \frac{\pi^2\hbar^2}{2ma^2} (1^2 + 1^2) = \frac{\pi^2\hbar^2}{2ma^2} (2) = \frac{\pi^2\hbar^2}{ma^2} $
- First Excited State: The next energy level corresponds to the first excited state. This occurs for the quantum number combinations $(n_x=1, n_y=2)$ and $(n_x=2, n_y=1)$. These states are degenerate (have the same energy). $ E_{1,2} = E_{2,1} = \frac{\pi^2\hbar^2}{2ma^2} (1^2 + 2^2) = \frac{\pi^2\hbar^2}{2ma^2} (1 + 4) = \frac{5\pi^2\hbar^2}{2ma^2} $
Final Energy Result
The energy of the first excited state is $\frac{5\pi^2\hbar^2}{2ma^2}$.
Important Questions from Schrödinger Equation 1D Potentials Harmonic Oscillator
- The energy $E$ and degeneracy $d$ of the second excited state of a three-dimensional, isotropic quantum harmonic oscillator with angular frequency $\omega$ are
- A particle of mass $ m $ is in a potential $ V(x) = \frac{1}{2}m\omega^2x^2 $ for $ x > 0 $ and $ V(x) = \infty $ for $ x \leq 0 $, where $ \omega $ is the angular frequency. The ratio of the energies corresponding to the lowest energy level to the next higher level is
- 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) ?
- Consider a particle in a two dimensional infinite square well potential of side $L$, with $0 \le x \le L$ and $0 \le y \le L$. The wavefunction of the particle is zero only along the line $y = \frac{L}{2}$, apart from the boundaries of the well. If the energy of the particle in this state is $E$, what is the energy of the ground state?
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 $\Psi(x, t)$ is
