Question

If the ground state energy of a particle in an infinite potential well of width L₁ is equal to the energy of the second excited state in another infinite potential well of width L₂, then the ratio \(\frac{L_1}{L_2}\) is equal to

• A. 1
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{\sqrt3}\)
• D. \(\frac{1}{9}\)

Answer 1

The Correct Option is B

To solve this problem, we need to understand the energy levels of a particle in an infinite potential well. The energy levels for a particle in an infinite potential well are given by the formula:

\(E_n = \frac{n^2 h^2}{8mL^2}\) 

where \(n\) is the quantum number (1, 2, 3,...), \(h\) is Planck's constant, \(m\) is the mass of the particle, and \(L\) is the width of the well.

1. For the first potential well (width \(L_1\)), the ground state energy (first energy level, \(n = 1\)) is:

\(E_1(L_1) = \frac{h^2}{8mL_1^2}\)

1. For the second potential well (width \(L_2\)), the energy of the second excited state (third energy level, \(n = 3\)) is:

\(E_3(L_2) = \frac{9h^2}{8mL_2^2}\)

1. According to the problem, the ground state energy of the first well is equal to the energy of the second excited state in the second well. Therefore:

\(\frac{h^2}{8mL_1^2} = \frac{9h^2}{8mL_2^2}\)

1. By simplifying the equation, the \(h^2\) and \((8m)\) terms cancel out. We get:

\(\frac{1}{L_1^2} = \frac{9}{L_2^2}\)

1. Taking square roots on both sides, we reach:

\(\frac{L_2}{L_1} = 3\)

1. Hence, the ratio \(\frac{L_1}{L_2}\) is:

\(\frac{L_1}{L_2} = \frac{1}{3}\)

Thus, the correct answer is \(\frac{1}{3}\).