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?
Show Hint
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.
- 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).
