Potential Barrier Wavefunction Analysis
We are considering an electron with energy $E$ approaching a potential barrier. The potential is defined as $V(x) = 0$ for $x < 0$ and $V(x) = V_0$ for $x > 0$. The condition given is $E < V_0$. We need to determine the form of the wavefunction $\psi(x)$ for $x > 0$.
Schrödinger Equation in the Barrier Region ($x > 0$)
For the region where $x > 0$, the potential is constant $V(x) = V_0$. The time-independent Schrödinger equation takes the form:
$ -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} + V_0\psi(x) = E\psi(x) $Rearranging this equation, we get:
$ \frac{d^2\psi(x)}{dx^2} = \frac{2m(V_0 - E)}{\hbar^2}\psi(x) $Wavefunction Form for $E < V_0$
Since $E < V_0$, the quantity $(V_0 - E)$ is positive. Let's define a positive real constant $\alpha$:
$ \alpha = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}} $Note that $\alpha$ is a real positive quantity as stated in the problem.
Substituting $\alpha^2 = \frac{2m(V_0 - E)}{\hbar^2}$ into the Schrödinger equation gives:
$ \frac{d^2\psi(x)}{dx^2} = \alpha^2\psi(x) $The general solution to this second-order differential equation is:
$ \psi(x) = A e^{\alpha x} + B e^{-\alpha x} $where $A$ and $B$ are constants.
Physically, for $x \to \infty$ (deep inside the barrier region), the wavefunction must be bounded. The term $A e^{\alpha x}$ increases exponentially as $x$ increases, which is not physically plausible as it implies infinite probability. Thus, we must set $A = 0$.
The physically valid wavefunction in the region $x > 0$ is therefore of the form $B e^{-\alpha x}$.
Final Solution
The space part of the wavefunction for $x > 0$, given $E < V_0$, must be of the form $e^{-\alpha x}$ (or proportional to it).
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) ?
