The wave function for a quantum mechanical particle in a 1-dimensional box of length 'a' is given by $\psi = A \sin \frac{\pi x}{a}$ The value of 'A' for a box of length 200 nm is
Particle in a Box: Normalization Constant Calculation
The normalization condition for a wave function $\psi(x)$ in a 1-dimensional box of length $a$ (from $x=0$ to $x=a$) is given by:
$ \int_0^a |\psi(x)|^2 dx = 1 $Given the wave function $\psi(x) = A \sin\left(\frac{\pi x}{a}\right)$, we substitute this into the normalization condition:
$ \int_0^a \left|A \sin\left(\frac{\pi x}{a}\right)\right|^2 dx = 1 $ $ A^2 \int_0^a \sin^2\left(\frac{\pi x}{a}\right) dx = 1 $To evaluate the integral, we use the trigonometric identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$. Here, $\theta = \frac{\pi x}{a}$.
$ \int_0^a \sin^2\left(\frac{\pi x}{a}\right) dx = \int_0^a \frac{1 - \cos\left(\frac{2\pi x}{a}\right)}{2} dx $ $ = \frac{1}{2} \left[ x - \frac{a}{2\pi} \sin\left(\frac{2\pi x}{a}\right) \right]_0^a $ $ = \frac{1}{2} \left[ \left(a - \frac{a}{2\pi} \sin(2\pi)\right) - \left(0 - \frac{a}{2\pi} \sin(0)\right) \right] $ $ = \frac{1}{2} [ (a - 0) - (0 - 0) ] = \frac{a}{2} $Now, substitute this result back into the equation for $A^2$:
$ A^2 \left( \frac{a}{2} \right) = 1 $ $ A^2 = \frac{2}{a} $ $ A = \sqrt{\frac{2}{a}} $Given the length of the box $a = 200$ nm:
$ A = \sqrt{\frac{2}{200 \text{ nm}}} $ $ A = \sqrt{\frac{1}{100 \text{ nm}}} $ $ A = \frac{1}{10 \text{ nm}^{1/2}} $ $ A = 0.1 \text{ nm}^{-1/2} $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 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)- 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 ________
