## To determine the total energy of a hydrogen atom in its excited state, we can use the formula for the total energy of an electron in a hydrogen atom:

E = -13.6 eV / n^2

where n is the principal quantum number representing the energy level. In this case, n = 6, so we can substitute it into the equation:

E = -13.6 eV / (6^2)

= -13.6 eV / 36

= -0.3778 eV

Therefore, the total energy of the hydrogen atom in its n = 6 excited state is approximately -0.3778 eV.

To determine the magnitude of the maximum angular momentum the electron can have, we can use the formula:

L = Ä§ * âˆš(l * (l + 1))

where Ä§ is the reduced Planck's constant (approximately 1.05457 x 10^-34 J s) and l is the orbital quantum number representing the shape of the orbit. In this case, since it is a hydrogen atom, l can range from 0 to (n-1). Therefore, l = (6-1) = 5.

Substituting the values into the equation:

L = 1.05457 x 10^-34 J s * âˆš(5 * (5 + 1))

â‰ˆ 9.433 J s

So, the magnitude of the maximum angular momentum the electron can have in this state is approximately 9.433 J s.

Lastly, the maximum value that the z component of the Lz of the angular momentum can have can be determined using the formula:

Lz = mÄ§

where m is the magnetic quantum number. For a given value of l, m can range from -l to +l. In this case, since l = 5, m can range from -5 to +5. Therefore, the maximum value of Lz would be obtained when m = 5.

Substituting the values into the equation:

Lz = 5 * 1.05457 x 10^-34 J s

â‰ˆ 5.273 J s

So, the maximum value that the z component of the Lz of the angular momentum can have is approximately 5.273 J s.