Question

What is the maximum possible angular momentum$L$(as a multiple of h) of a hydrogen atom with energy$-0.544eV$?

Short answer

The maximum possible angular momentum $L$is$\sqrt{20h}$.

Step-by-step solution

Given Information 

We need to find the maximum possible angular momentum$L$.

Simplify

Finding the maximum angular momentum $L$of a hydrogen atom.

Consider:

$E=-0.544eV$

know that the hydrogen atom's energy is given as:

$E_n=-\frac{13.6eV}{n^2}$

Using formula, find the value of $n$:

$$\begin{gathered} E_n=-\frac{13.6eV}{n^2} \\ -0.544eV=-\frac{13.6eV}{n^2} \\ \left(-0.544eV\right)\times n^2=-13.6eV \\ {n}^2=\frac{-13.6eV}{-0.544eV} \\ {n}^2=25 \\ n=\sqrt{25} \\ n=5 \end{gathered}$$

Now, finding the maximum angular momentum $L$.

Angular momentum $L$is given as:

$L=\sqrt{l\left(l+1\right)h}$

Then,

$$\begin{gathered} L=\sqrt{l\left(l+1\right)h} \\ L=\sqrt{4\left(4+1\right)h} \\ L=\sqrt{4\left(5\right)h} \\ L=\sqrt{20h} \end{gathered}$$