The wave function of a particle in a cubic box (of side L) is given by
$\psi(x, y, z) = \sqrt{32/L^3} \sin \frac{\pi x}{L} \cos \frac{\pi x}{L} \sin \frac{2\pi y}{L} \sin \frac{\pi z}{L}$.
The ratio of the energy of the state corresponding to the above wave function to the ground state energy is ________.
(rounded off to the nearest integer)
Wave Function Analysis
The given wave function is:
$ \psi(x, y, z) = \sqrt{\frac{32}{L^3}} \sin \frac{\pi x}{L} \cos \frac{\pi x}{L} \sin \frac{2\pi y}{L} \sin \frac{\pi z}{L} $
Using the double angle identity $ \sin(2\theta) = 2 \sin\theta \cos\theta $, we simplify $ \sin \frac{\pi x}{L} \cos \frac{\pi x}{L} $ to $ \frac{1}{2} \sin \frac{2\pi x}{L} $.
The wave function becomes:
$ \psi(x, y, z) = \sqrt{\frac{32}{L^3}} \left( \frac{1}{2} \sin \frac{2\pi x}{L} \right) \sin \frac{2\pi y}{L} \sin \frac{\pi z}{L} $
$ \psi(x, y, z) = \sqrt{\frac{8}{L^3}} \sin \frac{2\pi x}{L} \sin \frac{2\pi y}{L} \sin \frac{\pi z}{L} $
Quantum Numbers Identification
The standard wave function for a particle in a 3D cubic box is:
$ \psi_{n_x, n_y, n_z}(x, y, z) = \left(\frac{2}{L}\right)^{3/2} \sin \frac{n_x \pi x}{L} \sin \frac{n_y \pi y}{L} \sin \frac{n_z \pi z}{L} $
Comparing the simplified wave function, we find the quantum numbers $ (n_x, n_y, n_z) $:
- $ n_x = 2 $
- $ n_y = 2 $
- $ n_z = 1 $
The normalization constant $ \sqrt{8/L^3} $ is consistent with $ (2/L)^{3/2} $.
State Energy Calculation
The energy level formula for a particle in a 3D cubic box is:
$ E_{n_x, n_y, n_z} = \frac{\pi^2 \hbar^2}{2mL^2} (n_x^2 + n_y^2 + n_z^2) $
For the state $ (2, 2, 1) $, the energy $ E_{2,2,1} $ is:
$ E_{2,2,1} = \frac{\pi^2 \hbar^2}{2mL^2} (2^2 + 2^2 + 1^2) = \frac{\pi^2 \hbar^2}{2mL^2} (4 + 4 + 1) = \frac{9 \pi^2 \hbar^2}{2mL^2} $
Ground State Energy Calculation
The ground state has quantum numbers $ (1, 1, 1) $. The energy $ E_{1,1,1} $ is:
$ E_{1,1,1} = \frac{\pi^2 \hbar^2}{2mL^2} (1^2 + 1^2 + 1^2) = \frac{\pi^2 \hbar^2}{2mL^2} (3) = \frac{3 \pi^2 \hbar^2}{2mL^2} $
Energy Ratio Determination
The ratio of the state energy $ E_{2,2,1} $ to the ground state energy $ E_{1,1,1} $ is:
$ \text{Ratio} = \frac{E_{2,2,1}}{E_{1,1,1}} = \frac{\frac{9 \pi^2 \hbar^2}{2mL^2}}{\frac{3 \pi^2 \hbar^2}{2mL^2}} = \frac{9}{3} = 3 $
Rounded to the nearest integer, the ratio is 3.
Important Questions from Particle in a Box
Consider two non-interacting particles confined to a one-dimensional box with infinite potential barriers. Their wavefunctions are $\psi_1$ and $\psi_2$ and energies are $E_1$ and $E_2$, respectively. The INCORRECT statement(s) about this system is/are
- Wavefunctions and energies for a particle confined in a cubic box are $\psi_{n_x,n_y,n_z}$ and $E_{n_x,n_y,n_z}$, respectively. The functions $\Phi_1$, $\Phi_2$, $\Phi_3$, and $\Phi_4$ are written as linear combinations of $\psi_{n_x,n_y,n_z}$. Among these functions, the eigenfunction(s) of the Hamiltonian operator for this particle is/are
$\Phi_1 = \frac{1}{\sqrt{2}}\psi_{1,4,1} - \frac{1}{\sqrt{2}}\psi_{2,2,3}$
$\Phi_2 = \frac{1}{\sqrt{2}}\psi_{1,5,1} + \frac{1}{\sqrt{2}}\psi_{3,3,3}$
$\Phi_3 = \frac{1}{\sqrt{2}}\psi_{1,3,8} + \frac{1}{\sqrt{2}}\psi_{3,8,1}$
$\Phi_4 = \frac{1}{2}\psi_{3,3,1} + \frac{\sqrt{3}}{2}\psi_{2,4,1}$ - The wavelength associated with a particle in one-dimensional box of length $L$ is ($n$ refers to the quantum number)
- Assume 1,3,5-hexatriene to be a linear molecule and model the $\pi$ electrons as particles in a one-dimensional box of length 0.70 nm. The wavelength (in nm) corresponding to the transition from the ground-state to the first excited-state is ________
The $\pi$ electrons in benzene can be modelled as particles in a ring that follow Pauli's exclusion principle. Given that the radius of benzene is 1.4 Å, the longest wavelength of light that is absorbed during an electronic transition in benzene is ____________ nm. (Up to one decimal place. Use $m_e =9.1\times10^{-31} \text{ kg}$, $h=6.6\times10^{-34} \text{ Js}$, $c=3.0\times10^8 \text{ m s}^{-1}$)
