Question

A particle of mass \(m\) is confined in the ground state of a one-dimensional box extending from \(x = -2L\) to \(x = +2L\). The wave function of the particle in this state is \(\psi = \psi_0 \cos\dfrac{\pi x}{4L}\), where \(\psi_0\) is a constant. The energy eigenvalue corresponding to this state is:

• A. \(\hbar^2\pi^2 / 2mL^2\)
• B. \(\hbar^2\pi^2 / 32mL^2\)
• C. \(\hbar^2\pi^2 / 16mL^2\)
• D. \(\hbar^2\pi^2 / 4mL^2\)

Hint:

The well width is \(4L\). Use \(E_1 = \dfrac{\pi^2\hbar^2}{2m a^2}\) with \(a = 4L\).

Answer 1

The Correct Option is B

Step 1: Inside the box the potential is zero, so the wave function satisfies the free-particle time-independent Schrodinger equation:
\[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi. \]
Step 2: Write \(\psi = \psi_0\cos(kx)\) with \(k = \dfrac{\pi}{4L}\). Differentiating twice:
\[ \frac{d^2\psi}{dx^2} = -k^2\psi_0\cos(kx) = -k^2\psi. \]
Step 3: Substitute back:
\[ -\frac{\hbar^2}{2m}(-k^2\psi) = E\psi \;\Rightarrow\; E = \frac{\hbar^2 k^2}{2m}. \]
Step 4: Insert \(k = \dfrac{\pi}{4L}\):
\[ E = \frac{\hbar^2}{2m}\left(\frac{\pi}{4L}\right)^2 = \frac{\hbar^2\pi^2}{2m\cdot 16L^2} = \frac{\hbar^2\pi^2}{32mL^2}. \]
\[ \boxed{E = \dfrac{\hbar^2\pi^2}{32mL^2}} \]