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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,

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

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,

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

Putting n = 4 in the above equation, you get

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

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,

$L=\frac{nh}{2\mathrm{Ï€}}$

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

Substitute all known values in the above equation, you get

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

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