## To determine the total energy of a hydrogen atom in the n = 6 excited state, we can use the formula for the energy of a hydrogen atom at any given energy level:

En = -13.6 eV / n^2

where n is the principal quantum number. Plugging in n = 6 into the equation, we can calculate the total energy.

(a) The total energy = -13.6 eV / (6^2) = -13.6 eV / 36 = -0.3778 eV (approximately)

Now let's move on to the angular momentum of the electron. In quantum mechanics, the angular momentum is quantized and given by the formula:

L = sqrt(l(l+1) * h / (2π))

where l is the orbital quantum number and h is the Planck's constant.

For a hydrogen atom in the excited state with n = 6, the maximum possible value of l is n - 1. Therefore, l = 6 - 1 = 5.

(b) The magnitude of the maximum angular momentum = sqrt(5(5+1) * h / (2π)) = sqrt(30 * (6.626 × 10^-34 J s) / (2π)) = sqrt(9.822 × 10^-33 J s) (approximately)

Lastly, the maximum value that the z component (Lz) of the angular momentum can have is given by:

Lz = mℏ

where m is the magnetic quantum number and ℏ is the reduced Planck's constant (h / 2π).

For a given orbital quantum number l, the magnetic quantum number can take values ranging from -l to l.

Therefore, for l = 5,

(c) The maximum value that the z component (Lz) of the angular momentum can have = 5ℏ = 5 * (6.626 × 10^-34 J s) / (2π) (approximately)

Please note that the values provided here are approximate and may vary slightly due to rounding.