Chapter 29, Problem 22P

(a)

To determine

The possible values of $L$ for a hydrogen atom in $3d$ state.

Answer to Problem 22P

The possible values of $L$ for a hydrogen atom in $3d$ state is $2.58\times {10}^{-34}\text{J}\cdot \text{s}$.

Explanation of Solution

Write the expression for orbital angular momentum.

  $L=\sqrt{l\left(l+1\right)}\hslash$

Here, $L$ is the orbital angular momentum, $l$ is the orbital quantum number.

Since, the hydrogen is in $3d$ state; the principle quantum number will be $n=3$ and orbital quantum number is $l=2$.

Conclusion:

Substitute $2$ for $l$ and to find the orbital angular momentum

  $\begin{array}{c}L=\sqrt{2\left(2+1\right)}\\ =\sqrt{6}\hslash \end{array}$

Therefore, possible values of $L$ for a hydrogen atom in $3d$ state is $\sqrt{6}\hslash$.

(b)

To determine

The possible values of $L_z$ for a hydrogen atom in $3d$ state.

Answer to Problem 22P

The possible values of $L_z$ for a hydrogen atom in $3d$ state is $-2\hslash ,-\hslash ,0,\hslash \text{ and }2\hslash$.

Explanation of Solution

Write the expression for z-component of orbital angular momentum.

  $L_z=m_l\hslash$

Here, $L_z$ is the z-component of orbital angular momentum, $m_l$ is the magnetic quantum number.

Since, the hydrogen is in $3d$ state; the principle quantum number will be $n=3$. Thus, the magnetic quantum number can be $-2,-1,0,1,2$.

Conclusion:

Substitute $-2$ for $m_l$ to find the z-component orbital angular momentum for $m_l=-2$.

  $\begin{array}{c}{L}_z=\left(-2\right)\hslash \\ =-2\hslash \end{array}$

Substitute $-1$ for $m_l$ and to find the z-component orbital angular momentum for $m_l=-1$.

  $\begin{array}{c}{L}_z=\left(-1\right)\hslash \\ =-\hslash \end{array}$

Substitute $0$ for $m_l$ and to find the z-component orbital angular momentum for $m_l=0$.

  $\begin{array}{c}{L}_z=\left(0\right)\hslash \\ =0\end{array}$

Substitute $1$ for $m_l$ and to find the z-component orbital angular momentum for $m_l=1$.

  $\begin{array}{c}{L}_z=\left(1\right)\hslash \\ =\hslash \end{array}$

Substitute $2$ for $m_l$ to find the z-component orbital angular momentum for $m_l=2$.

  $\begin{array}{c}{L}_z=\left(2\right)\hslash \\ =2\hslash \end{array}$

Therefore, possible values of $L_z$ for a hydrogen atom in $3d$ state is $-2\hslash ,-\hslash ,0,\hslash \text{ and }2\hslash$.

(c)

To determine

The possible values of $\theta$ for a hydrogen atom in $3d$ state.

Answer to Problem 22P

The possible values of $\theta$ for a hydrogen atom in $3d$ state are $145°,114°,90.0°,65.9°,\text{ and }35.3°$.

Explanation of Solution

Write the expression for $\theta$.

  $\theta ={\cos }^{-1}\left(\frac{L_z}{L}\right)$

The possible values of $L_z$ for a hydrogen atom in $3d$ state is $-2\hslash ,-\hslash ,0,\hslash \text{ and }2\hslash$ and the possible values of $L$ for a hydrogen atom in $3d$ state is $2.58\times {10}^{-34}\text{J}\cdot \text{s}$.

Conclusion:

Substitute $-2\hslash$ for $L_z$, $2.58\times {10}^{-34}\text{J}\cdot \text{s}$ for $L$ and $1.05\times {10}^{-34}\text{J}\cdot \text{s}$ for $\hslash$ to find $\theta$.

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{-2\left(1.05\times {10}^{-34}\text{J}\cdot \text{s}\right)}{2.58\times {10}^{-34}\text{J}\cdot \text{s}}\right)\\ =\cos \left(-0.819\right)\\ =145°\end{array}$

Substitute $-\hslash$ for $L_z$, $2.58\times {10}^{-34}\text{J}\cdot \text{s}$ for $L$ and $1.05\times {10}^{-34}\text{J}\cdot \text{s}$ for $\hslash$ to find $\theta$.

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{\left(-1.05\times {10}^{-34}\text{J}\cdot \text{s}\right)}{2.58\times {10}^{-34}\text{J}\cdot \text{s}}\right)\\ ={\cos }^{-1}\left(-0.407\right)\\ =114°\end{array}$

Substitute $0$ for $L_z$, $2.58\times {10}^{-34}\text{J}\cdot \text{s}$ for $L$ and $1.05\times {10}^{-34}\text{J}\cdot \text{s}$ for $\hslash$ to find $\theta$.

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{0\left(1.05\times {10}^{-34}\text{J}\cdot \text{s}\right)}{2.58\times {10}^{-34}\text{J}\cdot \text{s}}\right)\\ ={\cos }^{-1}\left(0\right)\\ =90.0°\end{array}$

Substitute $\hslash$ for $L_z$, $2.58\times {10}^{-34}\text{J}\cdot \text{s}$ for $L$ and $1.05\times {10}^{-34}\text{J}\cdot \text{s}$ for $\hslash$ to find $\theta$.

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{1.05\times {10}^{-34}\text{J}\cdot \text{s}}{2.58\times {10}^{-34}\text{J}\cdot \text{s}}\right)\\ ={\cos }^{-1}\left(0.408\right)\\ =65.9°\end{array}$

Substitute $2\hslash$ for $L_z$, $2.58\times {10}^{-34}\text{J}\cdot \text{s}$ for $L$ and $1.05\times {10}^{-34}\text{J}\cdot \text{s}$ for $\hslash$ to find $\theta$.

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{2\left(1.05\times {10}^{-34}\text{J}\cdot \text{s}\right)}{2.58\times {10}^{-34}\text{J}\cdot \text{s}}\right)\\ ={\cos }^{-1}\left(0.816\right)\\ =35.3°\end{array}$

Therefore, possible values of $\theta$ for a hydrogen atom in $3d$ state are $145°,114°,90.0°,65.9°,\text{ and }35.3°$.