Question

Show that a transition where$\mathbf{∆}{\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{\pm }\mathbf{1}$corresponds to a dipole moment in the xy-plane, while $\mathbf{∆}{\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$ corresponds to a moment along the z-axis. (You need to consider only the $\mathit{\varphi }$ -parts of therole="math" localid="1659783155213" ${\mathit{\psi }}_i$ and${\mathit{\psi }}_{\mathbf{f}}$ , which are of the form ${\mathit{e}}^{\mathbf{i}{\mathbf{m}}_{\mathbf{l}}\mathbf{\varphi }}$):

Short answer

A transition where $∆m_l=\pm 1$ corresponds to a dipole moment in the XY-plane, while$∆m_l=0$ corresponds to a moment along the z-axis.

Step-by-step solution

Used Formula:

When two equal and opposite charges are separated by a distance in space, is called a dipole. The product of the distance between them and their magnitude is called the dipole moment of that dipole.

As you know that dipole moment will be

$$\begin{gathered} p=-eRe\left(e^{i∆Et/h}\int r{{\psi }^{*}}_f\left(\overline{r}\right){\psi }_i\left(\overline{r}\right)dV\right) \\ {{\psi }^{*}}_f\left(\overline{r}\right){\psi }_i\left(\overline{r}\right) \\ r=x\hat{x}+y\hat{y}+z\hat{z}......\left(1\right) \end{gathered}$$

By solving the product ${{\psi }^{*}}_f\left(\overline{r}\right){\psi }_i\left(\overline{r}\right)$and writing out the r vector in its Cartesian component, you get

$$\begin{gathered} r=x\hat{x}+y\hat{y}+z\hat{z} \\ r=r\sin \theta \cos \varphi \hat{x}+r\sin \theta \sin \varphi \hat{y}+r\cos \theta \hat{z}......\left(2\right) \end{gathered}$$

Now, by using equation (2) in equation (1) and leaving R(r) and general,

Where, r is the distance, $\theta$ is the colatitude, $\varphi$ is the azimuth, and $\psi$wave function.

Therefore,

$\int \left(r\sin \theta \cos \varphi \hat{x}+r\sin \theta \sin \varphi \hat{y}+r\cos \theta \hat{z}\right)=R_f\left(\overrightarrow{r}\right){⊝}_f\left(\theta \right)e^{-i{m}_l{l}_f}{R}_i\left(\overrightarrow{r}\right){⊝}_i\left(\theta \right)e^{-i{m}_l{l}_l}{r}^2\sin \theta drd\theta d\varphi$

x-component of equation (3):

The x-component involves the integral,

$l_x={\int }_0^{2x}\cos \varphi e^{+i(m{l}_l+m{l}_l)\varphi }d\varphi$

In the above integral, if$∆m_l=0$:

The term will be the cosine integral from 0 to $2\mathrm{\pi }$ which will give zero.

Where, is the magnetic quantum number.

If $∆m_l=\pm 1$

$$\begin{gathered} l_x={\int }_0^{2\mathrm{\pi }}\cos \varphi \left(\cos \varphi \pm i\sin \varphi \right)d\varphi \\ ={\int }_0^{2\mathrm{\pi }}{\cos }^2\varphi d\varphi \pm i{\int }_0^{2\mathrm{\pi }}\cos \varphi \sin \varphi d\varphi \\ ={\left[\frac{1}{4}\sin 2\varphi +\frac{\varphi }{2}\right]}_0^{2\mathrm{\pi }}+i{\left[-\frac{1}{2}{\cos }^2\varphi \right]}_0^{2\mathrm{\pi }} \\ =\left[\frac{1}{4}\sin \left(4\mathrm{\pi }\right)+\frac{2\mathrm{\pi }}{2}-\frac{1}{4}\sin \left(0\right)+0\right]+i\left[-\frac{1}{2}{\cos }^2\left(2\mathrm{\pi }\right)+\frac{1}{2}{\cos }^2\left(0\right)\right] \\ {l}_x=\left[\mathrm{\pi }\right]+i\left[-\frac{1}{2}+\frac{1}{2}\right] \\ =\mathrm{\pi } \end{gathered}$$

y-component of equation (3):

The y-component involves the integral,

$l_y={\int }_0^{2\mathrm{\pi }}\sin \varphi e^{+i{(m{l}_i+m{l}_f)}^{\varphi }}d\varphi$

In the above integral,

If $∆m_l=0$,

The term will be the sine integral from 0 to $2\mathrm{\pi }$ which will give zero.

If $∆m_l=\pm 1$,

$$\begin{gathered} l_y={\int }_0^{2\mathrm{\pi }}\sin \varphi \left(\cos \varphi \pm i\sin \varphi \right)d\varphi \\ ={\int }_0^{2\mathrm{\pi }}\cos \varphi \sin \varphi d\varphi \pm {\int }_0^{2\mathrm{\pi }}{\sin }^2\varphi d\varphi \\ =\pm \mathrm{\pi } \end{gathered}$$

z-component of equation (3):

The z-component involves the integral,

$l_z={\int }_0^{2\mathrm{\pi }}{e}^{+i\left(m{l}_i+m{l}_f\right)\varphi }d\varphi$

In the above integral,

If$∆m_l=0$,

The term will give$2\mathrm{\pi }$

If$∆m_l=\pm 1$,

It will give $\cos \varphi \pm i\sin \varphi$ integrated over one period, i.e., zero.

Conclusion:

From the steps 2, 3 and 4 given above, you get, $∆m_l=\pm 1$corresponds to a dipole moment in the xy-plane, while $∆m_l=0$corresponds to a moment along the z-axis.