Question: An electron is in an n = 4 state of the hydrogen atom. (a) What is its energy? (b) What properties besides energy are quantized, and what values might be found if these properties were to be measured?

The energy level of the hydrogen atom is, n = 4.

According to Bohr's atomic model, the electron is excited to a higher energy state by absorbing energy in the form of photons. At a higher energy level, the excited electron is less stable and quickly emits a photon to return to a lower, more stable energy level.

The emitted energy can be calculated using the following equation,

$E\left(n\right)=-\frac{1}{n^2}\times 13.6\text{eV}$

Here, n is the energy level of the electron.

The formula for the energy of an atom in the nth level of the hydrogen atom is given by,

$E\left(n\right)=-\frac{1}{n^2}\times 13.6\text{eV}$

Putting n = 4 in the above equation, you get

$$\begin{gathered} E\left(4\right)=-\frac{1}{{\left(4\right)}^2}\times 13.6\text{eV} \\ =-\frac{1}{16}\times 13.6\text{eV} \\ =-0.85\text{eV} \end{gathered}$$

Hence, the energy of the hydrogen atom in the given state is -0.85 eV.

The 'angular momentum magnitude' and the 'z-component of the angular momentum' are the two other properties that are quantized besides the energy.

The formula for the angular momentum of the hydrogen atom is given by,

$L=\frac{nh}{2\pi }$

Here, h is Planck’s constant and its value is $4.136\times {10}^{-15}\text{eV}·\text{s}$.

Substitute all known values in the above equation, you get

$$\begin{gathered} L=\frac{4\times 4.136\times {10}^{-15}\text{eV}·\text{s}}{2\pi } \\ =2.633\times {10}^{-15}\text{eV}·\text{s} \end{gathered}$$

Hence, the magnitude of the angular momentum of the hydrogen atom in the given state is$2.633\times {10}^{-15}\text{eV}·\text{s}$ .