Calculate the angular momentum of an electron in a hydrogen atom.

To calculate the angular momentum of an electron in a hydrogen atom, we can use the principles of quantum mechanics and specifically, the angular momentum operator. In quantum mechanics, angular momentum is quantized, meaning it can only take on certain discrete values. The angular momentum of an electron in a hydrogen atom is given by:

\[L = \sqrt{l(l+1)}ħ\]

Where:
- \(L\) represents the magnitude of the angular momentum.
- \(l\) is the orbital quantum number, which is a non-negative integer. It determines the shape of the electron's orbital. For hydrogen's ground state, \(l = 0\).
- \(ħ\) is the reduced Planck constant, approximately equal to \(1.0545718 × 10^{-34}\) Joule-seconds.

For the ground state of hydrogen (\(n = 1\), \(l = 0\)), the angular momentum is:

\[L = \sqrt{0(0+1)}ħ = 0ħ = 0\]

So, for the ground state of a hydrogen atom, the angular momentum of the electron is zero. This means that the electron in the hydrogen atom's lowest energy state, often referred to as the \(1s\) orbital, has no orbital angular momentum. However, it still possesses intrinsic angular momentum, known as spin, which is a quantum property intrinsic to particles like electrons.

Keep in mind that for higher energy states of the hydrogen atom, where \(n > 1\), the angular momentum values will be different and quantized according to the above formula, reflecting the various possible electron orbital shapes and sizes in those states.