91影视

The potential energy between the bound atoms,$U\left(r\right)=\frac{1}{2}k{x}^2$.

x is the deviation of the atomic separation from its equilibrium value,

The reduce mass is $渭$.

The equation 7.30 is given by,

$-\frac{{魔}^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)+\frac{{魔}^{2}I\left(I+1\right)}{2m{r}^2}R\left(r\right)+U\left(r\right)R\left(r\right)-ER\left(r\right)=0$

(a)

Consider the equation 7.30 as,

$-\frac{{魔}^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)+\frac{{魔}^{2}I\left(I+1\right)}{2m{r}^2}R\left(r\right)+U\left(r\right)R\left(r\right)-ER\left(r\right)=0$

Substitute$\frac{1}{2}k{x}^2$for U(r) and$渭$ form into above equation.

$-\frac{{鈩�}^2}{2渭}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)+\frac{{鈩�}^{2}l(l+1)}{2渭r^2}R\left(r\right)+\frac{1}{2}k{x}^{2}R\left(r\right)-ER\left(r\right)=0$ 鈥︹�� (1)

According to the definition,

$f\left(r\right)=rR\left(r\right)$

Differentiate the above equation with respect to r.

$\frac{d}{dr}f\left(r\right)=R\left(r\right)+r\frac{dR\left(r\right)}{dr}$

Differentiate the above equation with respect to r.

$\begin{array}{r}\frac{d^{2}f\left(r\right)}{d{r}^2}=\frac{dR\left(r\right)}{dr}+r\frac{d^{2}R\left(r\right)}{d{r}^2}+\frac{dR\left(r\right)}{dr}\\ \frac{d^{2}f\left(r\right)}{d{r}^2}=2\frac{dR\left(r\right)}{dr}+r\frac{d^{2}R\left(r\right)}{d{r}^2}\\ \frac{d^{2}f\left(r\right)}{d{r}^2}=\frac{1}{r}\left(2r\frac{dR\left(r\right)}{dr}+r^2\frac{d^{2}R\left(r\right)}{d{r}^2}\right)\end{array}$

The term$\frac{1}{r}\left(2r\frac{dR\left(r\right)}{dr}+r^2\frac{d^{2}R\left(r\right)}{d{r}^2}\right)$in the above equation can be written as $\frac{d}{dr}\left(r^2\frac{d^{2}R\left(r\right)}{d{r}^2}\right)$.

$$\begin{gathered} \frac{d^{2}f\left(r\right)}{d{r}^2}=\frac{d}{dr}\left(r^2\frac{d^{2}R\left(r\right)}{d{r}^2}\right) \\ \frac{d^{2}f\left(r\right)}{d{r}^2}=\frac{1}{r}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right) \end{gathered}$$

Substitute$\frac{d^{2}f\left(r\right)}{d{r}^2}$for$\frac{1}{r}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)$into equation (1).

$-\frac{{鈩�}^2}{2渭}\frac{1}{r}\frac{d^{2}f\left(r\right)}{d{r}^2}\frac{{鈩�}^{2}l(l+1)}{2渭r^2}R\left(r\right)+\frac{1}{2}k{x}^{2}R\left(r\right)-ER\left(r\right)=0$

Substitute $f\left(r\right)/r$ for R(r) into above equation.

role="math" localid="1660026902814" $$\begin{gathered} 鈭�\frac{{魔}^2}{2渭}\frac{1}{r}\frac{d^{2}f\left(r\right)}{d{r}^2}+\frac{{魔}^{2}I(I+1)}{2渭r^2}\frac{f\left(r\right)}{r}+\frac{1}{2}k{x}^2\frac{f\left(r\right)}{r}鈭�E\frac{f\left(r\right)}{r}=0 \\ 鈭�\frac{{魔}^2}{2渭}\frac{1}{r}\frac{d^{2}f\left(r\right)}{d{r}^2}+\frac{{魔}^{2}I(I+1)}{2渭r^2}\frac{f\left(r\right)}{r}+\frac{1}{2}k{x}^2\frac{f\left(r\right)}{r}=E\frac{f\left(r\right)}{r}=0 \\ 鈭�\frac{{魔}^2}{2渭}\frac{d^{2}f\left(r\right)}{d{r}^2}+\frac{{魔}^{2}I(I+1)}{2渭r^2}f\left(r\right)+\frac{1}{2}k{x}^{2}f\left(r\right)=Ef\left(r\right).......\left(2\right) \end{gathered}$$

Hence the equation$鈭�\frac{{魔}^2}{2渭}\frac{d^{2}f\left(r\right)}{d{r}^2}+\frac{{魔}^{2}I(I+1)}{2渭r^2}f\left(r\right)+\frac{1}{2}k{x}^{2}f\left(r\right)=Ef\left(r\right)$is obtained for $f\left(r\right)=rR\left(r\right)$.

Now consider the coordinate transformation $x=r-a$.

$$\begin{gathered} x=r-a \\ r=x+a \end{gathered}$$

Substitute$x+a$ forr and$g\left(x\right)$ for$f\left(r\right)$ into equation (2).

$-\frac{{鈩�}^2}{2渭}\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{{鈩�}^{2}l(l+1)}{2渭(x+a{)}^2}g\left(x\right)+\frac{1}{2}k{x}^{2}g\left(x\right)=Eg\left(x\right)$ 鈥︹�� (3)

Hence the required equation$-\frac{{鈩�}^2}{2渭}\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{{鈩�}^{2}l(l+1)}{2渭(x+a{)}^2}g\left(x\right)+\frac{1}{2}k{x}^{2}g\left(x\right)=Eg\left(x\right)$ is obtained for $f\left(r\right)=g\left(x\right)$.

(b)

It is given that x<<a. Therefore,${\left(x+a\right)}^2鈮�a^2$.

Substitute$a^2$for${\left(x+a\right)}^2$into equation (3).

$$\begin{gathered} 鈭�\frac{{魔}^2}{2渭}\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{{魔}^{2}I(I+1)}{2渭a^2}g\left(x\right)+\frac{1}{2}k{x}^{2}g\left(x\right)=Eg\left(x\right) \\ 鈭�\frac{{魔}^2}{2渭}\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{1}{2}k{x}^{2}g\left(x\right)=Eg\left(x\right)鈭�\frac{{魔}^{2}I(I+1)}{2渭a^2}g\left(x\right) \\ 鈭�\frac{{魔}^2}{2渭}\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{1}{2}k{x}^{2}g\left(x\right)=\left[E鈭�\frac{{魔}^{2}I(I+1)}{2渭a^2}\right]g\left(x\right) \end{gathered}$$ 鈥︹�� (4)

Compare the equation (5) with 5.25 of the textbook equation, which describe the simple harmonic motion with mass $渭$, oscillation frequency ${蝇}_0=\sqrt{\frac{k}{渭}}$and effective energy $E^{\text{'}}=E-\frac{{魔}^{2}l\left(l+1\right)}{2渭a^2}$.

According to the simple harmonic oscillation given in section 5.7 of the textbook, the equation (4) can be written as,

$E^{\text{'}}=\left(n+\frac{1}{2}\right)魔{蝇}_0$

Substitute $\sqrt{\frac{k}{渭}}$for ${蝇}_0$into above equation.

$E^{\text{'}}=\left(n+\frac{1}{2}\right)魔\sqrt{\frac{k}{渭}}$

Therefore, energy for the diatomic molecules is given by,

$E=(\mathrm{n}+\frac{1}{2})魔\sqrt{\frac{\mathrm{k}}{\mathrm{渭}}}+\frac{{\mathrm{魔}}^2\mathrm{I}(\mathrm{I}+1)}{2{\mathrm{渭补}}^2}\begin{array}{c}\mathrm{n}=0,1,2,...\\ \mathrm{I}=0,1,2,...\end{array}$

Hence, the required equation is obtained for x<<a.

(c)

Therefore, energy for the diatomic molecules is given by,

$E=(\mathrm{n}+\frac{1}{2})魔\sqrt{\frac{\mathrm{k}}{\mathrm{渭}}}+\frac{{\mathrm{魔}}^2\mathrm{I}(\mathrm{I}+1)}{2{\mathrm{渭补}}^2}\begin{array}{c}\mathrm{n}=0,1,2,...\\ \mathrm{I}=0,1,2,...\end{array}$

The above equation contains the two terms, first term represents the vibrational energy at the frequency ${蝇}_0$. This term is associated with the harmonic oscillation during the separation of the diatomic molecules. The second term represent the rotational energy with angular momentum$L=魔\sqrt{I\left(I+1\right)}$ and moment of inertia $渭a^2$.This energy is associated with the rotation of the mass about an axis of the center of the mass at the distance .