You put 5 non-interacting identical fermions each of mass m into a 1-d box of dimension L.You then put 10 non-interacting bosons each of mass m into a 1-d box of length 2L.Which system has the lowest ground-state energy and what is the value of the fermion system ground-state energy divided by the boson system ground-state energy?

Question

Quizplus · Accepted Answer

For the fermion system, we know that each fermion can only occupy a single energy level due to the Pauli exclusion principle. The lowest energy state for this system is when all 5 fermions are each in the lowest energy level (n=1) such that the total energy is given by:

E_fermion = 5*(h_bar^2*pi^2)/(2*m*L^2)

For the boson system, there are no restrictions on the number of bosons in a single energy level. The lowest energy state for this system is when all 10 bosons are in the lowest energy level (n=1) such that the total energy is given by:

E_boson = 10*(h_bar^2*pi^2)/(8*m*L^2)

Dividing the fermion system energy by the boson system energy, we get:

E_fermion/E_boson = 5/8 * 8/10 = 5/10 = 1/2

Therefore, the correct answer is C because the boson system has the lowest ground-state energy and the fermion system ground-state energy is 2 times higher than the boson system ground-state energy.

You Put 5 Non-Interacting Identical Fermions Each of Mass M