Energy is proportional to $n_x^2+n_y^2+n_z^2$, with each quantum number a positive integer. The first distinct sums are 3, 6, 9, 11, 12 and 14. Thus the fifth excited level has sum 14, from $(1,2,3)$ and its permutations. All three numbers differ, so there are $3!=6$ states. The particle mass and well length scale energies but do not change this degeneracy.
More One dimensional potentials: step potential, finite rectangular well, tunnelling from a potential barrier, particle in 1,2,3-dimensional box, particle in single and double delta function potentials, 1,2,3 dimensional harmonic oscillator: concept of degeneracy questions
GATE 2024 PH Q13– A particle in an infinite one-dimensional well has (x,t)=23e-iE 1t/ 1(x)+16ei/6e-iE 2t/…
GATE 2024 PH Q22– A particle has V(x)= for x0, V(x)=V 0 for a x b, and zero elsewhere, with b>a>0. For E<V…
GATE 2024 PH Q47– V(x)=V 0+12m 02x2 for x > 0 and infinity for x ≤ 0. Figures P,Q,R,S specify offsets and…
GATE 2025 PH Q13– A particle of mass m is in a potential V(x)=12m2x2 for x>0 and V(x)= for x0, where is the…
GATE 2025 PH Q43– Let m, n denote energy eigenstates of a one-dimensional simple harmonic oscillator, with…
GATE 2026 PH Q39– The Hamiltonian for a quantum particle of mass m is given below, where <. The Schrödinger…