# Problem: The orbital angular momentum of an electron has a magnitude of 4.716 × 10–34 kg•m2/s. What is the angular-momentum quantum number l for this electron?

###### FREE Expert Solution

The orbital angular momentum, L:

$\overline{){\mathbf{L}}{\mathbf{=}}\sqrt{\mathbf{l}\mathbf{\left(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{\right)}}{\mathbf{\hslash }}}$, where ℏ = h/2π and h is the Plank's constant.

Solving for l:

$\begin{array}{rcl}{\mathbf{L}}^{\mathbf{2}}& \mathbf{=}& \mathbf{l}\mathbf{\left(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{\right)}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi }}^{\mathbf{2}}}\\ \mathbf{l}\mathbf{\left(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{\right)}& \mathbf{=}& \frac{{\mathbf{L}}^{\mathbf{2}}\mathbf{4}{\mathbf{\pi }}^{\mathbf{2}}}{{\mathbf{h}}^{\mathbf{2}}}\\ {\mathbf{l}}^{\mathbf{2}}\mathbf{+}\mathbf{l}& \mathbf{=}& \frac{{\mathbf{\left(}\mathbf{4}\mathbf{.}\mathbf{716}\mathbf{×}\mathbf{10}\mathbf{-}\mathbf{34}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{\left(}\mathbf{\pi }{\mathbf{\right)}}^{2}}{{\mathbf{\left(}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{×}\mathbf{10}\mathbf{-}\mathbf{34}\mathbf{\right)}}^{\mathbf{2}}}\end{array}$

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###### Problem Details

The orbital angular momentum of an electron has a magnitude of 4.716 × 10–34 kg•m2/s. What is the angular-momentum quantum number l for this electron?