Question

For a quantum particle confined inside a cubic box of side \( L \), the ground state energy is given by \( E_0 \). The energy of the first excited state is

• A. \( 2E_0 \)
• B. \( \sqrt{2}E_0 \)
• C. \( 3E_0 \)
• D. \( 6E_0 \)

Hint:

The energy levels for a particle in a box are quantized and proportional to \( n^2 \). The ground state corresponds to \( n = 1 \), and the first excited state corresponds to \( n = 2 \).

Answer 1

The Correct Option is A

Step 1: Energy levels of a particle in a box.
For a particle in a cubic box, the energy levels are quantized. The ground state energy is \( E_0 \), and the energy levels are given by: \[ E_n = \frac{n^2 h^2}{8mL^2} \] where \( n \) is a positive integer, \( h \) is Planck's constant, \( m \) is the particle's mass, and \( L \) is the length of the box.
Step 2: Finding the energy of the first excited state.
The ground state energy corresponds to \( n = 1 \). The first excited state corresponds to the next value of \( n = 2 \). So, the energy of the first excited state is: \[ E_2 = \frac{4h^2}{8mL^2} = 4E_0 \]
Step 3: Conclusion.
Thus, the energy of the first excited state is \( 3E_0 \), so the correct answer is (C).