Question:

Consider a particle in a two dimensional infinite square well potential of side $L$, with $0 \leq x \leq L$ and $0 \leq y \leq L$. The wavefunction of the particle is zero only along the line $y = \dfrac{L}{2}$, apart from the boundaries of the well. If the energy of the particle in this state is $E$, what is the energy of the ground state?

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In a two-dimensional infinite square well, the energy levels are determined by the quantum numbers in both directions. The ground state corresponds to the lowest quantum numbers for both directions.
Updated On: Aug 30, 2025
  • $\dfrac{1}{4}E$
  • $\dfrac{2}{5}E$
  • $\dfrac{3}{8}E$
  • $\dfrac{1}{2}E$
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The Correct Option is B

Solution and Explanation

- In a two-dimensional infinite square well, the energy levels are quantized. The energy of the particle is given by: \[ E_{n_x, n_y} = \dfrac{n_x^2\pi^2 \hbar^2}{2mL^2} + \dfrac{n_y^2\pi^2 \hbar^2}{2mL^2}, \] where $n_x$ and $n_y$ are the quantum numbers for the $x$ and $y$ directions.
- The given state has the wavefunction zero along $y = \dfrac{L}{2}$, which means the particle is constrained to half of the well in the $y$ direction. The energy in this state is not the ground state but is higher than the ground state.
- The ground state corresponds to $n_x = 1$ and $n_y = 1$, so the energy of the ground state is the lowest possible energy. The ratio of the ground state energy to the given state energy is $\dfrac{2}{5}E$.
Thus, the correct answer is (B).
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