91Ó°ÊÓ

An electron in a hydrogen atom is in a state with $\mathcal{l}=5$.

Using the concept of orbital angular momentum and its component value, we can get the required semi-classical angle between the two vectors substituting the value of l = 5 in the formula.

Formulae:

$\stackrel{→}{L}=\sout{h}\sqrt{\mathcal{l}\left(\mathcal{l}+1\right)}$ …...(¾±)

• The magnitude of the z-component of the angular momentum, $\stackrel{→}{L_z}=m_{\mathcal{l}}\sout{h}\mathrm{where}{m}_{\mathcal{l}}=-\mathcal{l}\mathrm{to}+\mathcal{l}$ …â¶Ä¦.(¾±¾±)
• The semi-classical angle between a vector and its component,

$\mathrm{θ}={\cos }^{-1}\left(\frac{\stackrel{→}{L_z}}{\stackrel{→}{L}}\right)$ … (iii)

Using the concept, we can say that the smallest possible value of the angle is achieved when$m_{\mathcal{l}}=\mathcal{l}$.

Thus, using the formulae of equations (i) and (ii) in equation (iii) with the given value$\mathcal{l}=5$, we can get the value of the semi-classical angle between$\stackrel{→}{L}$and $\stackrel{→}{L_z}$as follows:

$$\begin{gathered} \mathrm{θ}={\cos }^{-1}\left(\frac{\mathcal{l}\sout{h}}{\sqrt{\mathcal{l}\left(\mathcal{l}+1\right)\sout{h}}}\right) \\ ={\cos }^{-1}\left(\frac{5}{\sqrt{5\left(5+1\right)}}\right) \\ ={\cos }^{-1}\left(\frac{5}{\sqrt{30}}\right) \\ =24.1° \end{gathered}$$

Hence, the value of the angle is $24.1°$.