Derive the expression for the energy levels of a particle in a one-dimensional box of length L.
Understand the Problem
The question is asking to derive the expression for the energy levels of a quantum particle confined in a one-dimensional box of length L, which involves using quantum mechanics principles and boundary conditions.
Answer
The energy levels of a quantum particle in a one-dimensional box are given by $$ E_n = \frac{n^2\hbar^2\pi^2}{2mL^2} $$
Answer for screen readers
The expression for the energy levels of a quantum particle in a one-dimensional box is
$$ E_n = \frac{n^2\hbar^2\pi^2}{2mL^2} $$
where $n$ is a positive integer.
Steps to Solve
- Write the Time-Independent Schrödinger Equation
We start with the time-independent Schrödinger equation for a particle in one dimension:
$$ -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} = E\psi(x) $$
where $\hbar$ is the reduced Planck's constant, $m$ is the mass of the particle, $E$ is the energy, and $\psi(x)$ is the wave function.
- Set Up the Boundary Conditions
For a particle in a one-dimensional box of length $L$, the wave function must satisfy the following boundary conditions:
$$ \psi(0) = 0 \quad \text{and} \quad \psi(L) = 0 $$
This means the wave function must be zero at the ends of the box.
- Solve the Differential Equation
We solve the differential equation by rearranging it:
$$ \frac{d^2\psi(x)}{dx^2} = -\frac{2mE}{\hbar^2} \psi(x) $$
Letting
$$ k^2 = \frac{2mE}{\hbar^2} $$
the general solution to this equation is:
$$ \psi(x) = A\sin(kx) + B\cos(kx) $$
- Apply Boundary Conditions
Now we apply the boundary condition $\psi(0) = 0$. This gives us:
$$ \psi(0) = A\sin(0) + B\cos(0) = B = 0 $$
Thus,
$$ \psi(x) = A\sin(kx) $$
Now apply the second boundary condition $\psi(L) = 0$:
$$ \psi(L) = A\sin(kL) = 0 $$
For this to hold, we need
$$ kL = n\pi $$
where $n$ is a positive integer ($n = 1, 2, 3, \ldots$).
- Find the Expression for k and Energy Levels
Solving for $k$ gives us:
$$ k = \frac{n\pi}{L} $$
Now, substitute $k$ back into the expression for energy:
$$ E = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2 $$
This simplifies to:
$$ E_n = \frac{n^2\hbar^2\pi^2}{2mL^2} $$
- Final Result
Thus, the expression for the energy levels of a quantum particle in a one-dimensional box is given by:
$$ E_n = \frac{n^2\hbar^2\pi^2}{2mL^2} $$
The expression for the energy levels of a quantum particle in a one-dimensional box is
$$ E_n = \frac{n^2\hbar^2\pi^2}{2mL^2} $$
where $n$ is a positive integer.
More Information
This result shows that the energy levels are quantized, meaning a particle can only have specific energy values determined by the quantum number $n$. The energy increases with the square of this quantum number, indicating that higher energy states are further apart in terms of energy.
Tips
- Neglecting Boundary Conditions: One common mistake is not applying the boundary conditions correctly, which can lead to incorrect forms of the wave function. Always ensure that the wave function is zero at the boundaries.
- Forgetting Quantization: Sometimes, students may forget that $n$ must be a positive integer, which can lead to non-physical solutions.
