A particle of mass m is confined in a two-dimensional box that has sides L_x = L and L_y = 2L.By what factor is the energy of the 3rd excited state larger than the energy of the ground state?

Question

Quizplus · Accepted Answer

The energy of the particle in a two-dimensional box is given by:

E = (h^2 / 8m) * (n1^2 / L^2 + n2^2 / (2L)^2)

where n1 and n2 are the quantum numbers along the x and y directions, respectively.

For the ground state, n1=1 and n2=1, so the energy is:

E1 = (h^2 / 8m) * (1/L^2 + 1/(2L)^2) = (5/8) * (h^2 / 2mL^2)

For the 3rd excited state, n1=3 and n2=1, so the energy is:

E2 = (h^2 / 8m) * (9/L^2 + 1/(2L)^2) = (17/8) * (h^2 / 2mL^2)

The factor by which E2 is larger than E1 is then:

E2/E1 = (17/8) / (5/8) = 17/5

Therefore, the answer is D) 17/5.

A Particle of Mass M Is Confined in a Two-Dimensional