Question

detailed solutions/equations for both***

Part 1

Five electrons are in a two-dimensional square potential energy well with sides of length L.The potential energy is infinite at the sides and zero inside. The single particle energies are given by (h2/8mL2)(n2x +n2y), where nx and ny are integers. In units of (h2/8mL2) the energy of the ground state of the system

Part2

Electrons are in a two-dimensional square potential energy well with sides of length L. The potential energy is infinite at the sides and zero inside. The single-particle energies are given by (h2/8mL2)(n2x + n2y), where nx and ny are integers. At most the number of electrons that can have energy 8(h2/8mL2) is:

Answer 1

Part 1:

    The energy of particle in two dimensional square potential well

                           E = (h²/ 8mL²) ( n_x² + n_y²)

Five electrons are inside the well.

            here every state has occupied by 2 electons

               electron occupied the state, 1_x - 2 electron,   1_y - 2 electron

                                                         2_x - 1 electron

                                 E = (h²/ 8mL²)( 1² + 1² + 2²)

                                     E = 6 (h² / 8mL²)

                 The energy of ground state of the five electron system is 6 (h²/ 8mL²)

Part 2:

                  Energy of the system   E= 8 (h²/8mL²)

                     here n² = 8

   Possible states are n_x = 2, n_y= 2

   n² = 2² + 2² =8

State equavalent to the energy E= 8 (h²/ 8mL²) is   E_x,y = E_2,2

          Every state is occupied by two electrons 2_x = 2 electron,

                                                                       2_y=2 electron

                   The number of electron occupied the state E_2,2 = 4 electrons