Solution:

Given: Energy of particle = 66h^2/8mL^2

To find: Degeneracy of the particle

Formula used: The energy of a particle in a 3D box is given by

E = [(n1^2 + n2^2 + n3^2) x h^2]/8mL^2

where n1, n2, and n3 are the three quantum numbers.

Explanation:

Degeneracy is the number of ways in which a particular energy level can be attained by a system. In other words, it is the number of quantum states that correspond to a given energy level.

In our case, we have been given the energy of a particle in a 3D box. The energy of a particle in a 3D box can be expressed in terms of the three quantum numbers as follows:

E = [(n1^2 + n2^2 + n3^2) x h^2]/8mL^2

where n1, n2, and n3 are the three quantum numbers.

To find the degeneracy, we need to find the number of ways in which the three quantum numbers can be combined to give the same energy level. For this, we can use the following formula:

Degeneracy = (1/6) x [(n1 + n2 + n3)^2 + 3(n1^2 + n2^2 + n3^2)]

Substituting the given values, we get:

E = [(n1^2 + n2^2 + n3^2) x h^2]/8mL^2

66h^2/8mL^2 = [(n1^2 + n2^2 + n3^2) x h^2]/8mL^2

n1^2 + n2^2 + n3^2 = 66

Using this value of n1^2 + n2^2 + n3^2 in the formula for degeneracy, we get:

Degeneracy = (1/6) x [(n1 + n2 + n3)^2 + 3(n1^2 + n2^2 + n3^2)]

Degeneracy = (1/6) x [(n1 + n2 + n3)^2 + 3(66)]

Now, we need to find the values of n1, n2, and n3 that satisfy the condition n1^2 + n2^2 + n3^2 = 66.

We can do this by using the following table:

n1 n2 n3 n1^2 + n2^2 + n3^2
0 0 8 64
0 2 6 40
0 4 4 32
2 2 4 24
2 4 2 24
4 4 0 32
Total 216

We can see that there are 6 possible combinations of n1, n2, and n3 that satisfy the condition n1^2 + n2^2 + n3^2 = 66.

Therefore, the degeneracy of the particle is 6 x (1/6) x