Effect of Coulomb term on optical properties of fluorine

Abstract

In this work, the optical properties of fluorine have been theoretically investigated. The Woods–Saxon potential model has been considered together with the spin–orbit interaction term. Then, the Schrödinger equation is solved by employing the Nikiforov-Uvarov method without and with considering the Coulomb term. We determine refractive index changes (RIC) and absorption coefficient (AC) of the fluorine with and without considering the Coulomb term. We found that (i) the RIC increases and shifts toward higher energies considering the Coulomb term. (ii) AC raises and moves to higher energies considering the Coulomb term. We also deduced that the Coulomb term has an important effect on the optical properties of fluorine.

Appendices

Appendix A

In this part, we have presented the NU method. The procedure as a powerful mathematical tool is employed to solve the second-order differential equations using the special orthogonal functions (Turkoglu et al. 2021). Consider the below equation

$$\frac{{d}^{2}\Phi \left(z\right)}{d{z}^{2}}+\frac{\lambda \left(z\right)}{\eta \left(z\right)}\frac{d\Phi \left(z\right)}{dz}+\frac{\gamma \left(z\right)}{{\eta }^{2}\left(z\right)}\Phi \left(z\right)=0,$$

(17)

where $\eta \left(z\right)$ and $\gamma \left(z\right)$ are second-order polynomials and $\lambda \left(z\right)$ is a first-order polynomial. We use the following function.

$$\Phi \left(z\right)=\varphi \left(z\right){y}_{n}\left(z\right)$$

(18)

Therefore, we obtain

$$\eta \left(z\right)\frac{{d}^{2}y\left(z\right)}{d{z}^{2}}+\mu \left(z\right)\frac{dy\left(z\right)}{dz}+\delta y\left(z\right)=0$$

(19)

and

$$\frac{\frac{d\varphi (z)}{dz}}{\varphi (z)}=\frac{\mu (z)}{\eta (z)}$$

(20)

The function $y(z)$ is of the hypergeometric type. It can be written as (Ahn and Chuang 1987)

$${y}_{n}\left(z\right)=\frac{{\alpha }_{n}}{\theta (z)}\frac{{d}^{n}}{d{z}^{n}}\left[{\eta }^{n}(z)\theta (z)\right]$$

(21)

where ${\alpha }_{n}$ is the normalization constant. To solve Eq. (20), the following relations should be satisfied

$$\mu \left(z\right)=\lambda \left(z\right)+2\omega \left(z\right), \frac{d}{dz}\left[\theta (z)\eta (z)\right]=\mu \left(z\right)\theta \left(z\right)$$

(22)

$$\omega \left( z \right) = \frac{1}{2}\left[ {\frac{d\eta \left( z \right)}{{dz}} - \lambda \left( z \right)} \right] \pm \sqrt {\frac{{\left[ {\frac{d\eta \left( z \right)}{{dz}} - \lambda \left( z \right)} \right]^{2} }}{4} - \gamma \left( z \right) + k\eta \left( z \right)}$$

(23)

$$\delta = k + \omega^{\prime }$$

(24)

$$\delta = \delta_{n} = - n\mu^{\prime } - \frac{{n\left( {n + 1} \right)}}{2}\eta^{\prime \prime } ,n = 0, 1, 2, \ldots$$

(25)

Appendix B

To solve Eq. (4), we used the following approximation

$$\frac{l\left(l+1\right)}{{r}^{2}}\approx \frac{l\left(l+1\right)}{{r}_{m}^{2}}\left[{B}_{0}+{B}_{1}\left(\frac{-{e}^{-\rho r}}{1+p{e}^{-\rho r}}\right)+{B}_{2}{\left(\frac{-{e}^{-\rho r}}{1+p{e}^{-\rho r}}\right)}^{2}\right]$$

(26)

where ${r}_{m}$ is a minimum point, $\rho =\frac{1}{a}$, and $p={e}^{\rho {R}_{0}}$. Now, we employ the new variable as $z=-{e}^{-\rho r}$. Thus, we can express Eq. (4) as follows.

$$z^{2} \frac{{d^{2} R\left( z \right)}}{{dz^{2} }} + z\frac{dR\left( z \right)}{{dz}} + \frac{2m}{{\hbar^{2} }}\left[ {E - V_{0} p\left( {\frac{z}{1 - pz}} \right) + \frac{{0.44r_{0}^{2} V_{0} p}}{{2ar_{0}^{\prime 2} }}\left( {j\left( {j + 1} \right) - l\left( {l + 1} \right) - \frac{3}{4}} \right)\left( {\frac{z}{1 - pz}} \right)^{2} - \frac{{\hbar^{2} }}{2m}\frac{{l\left( {l + 1} \right)}}{{r_{m}^{2} }}\left( {B_{0} + B_{1} \left( {\frac{{ - e^{ - \rho r} }}{{1 + pe^{ - \rho r} }}} \right) + B_{2} \left( {\frac{{ - e^{ - \rho r} }}{{1 + pe^{ - \rho r} }}} \right)^{2} } \right)} \right]R\left( z \right) = 0$$

(27)

We define the three following variables

$$\varepsilon =\sqrt{\frac{l(l+1){a}^{2}}{{r}_{m}^{2}}{B}_{0}-\frac{2mE{a}^{2}}{{\hslash }^{2}}}$$

(28)

$${\zeta }_{1}=2p{\varepsilon }^{2}-\frac{2m{V}_{0}p{a}^{2}}{{\hslash }^{2}}-\frac{l\left(l+1\right){a}^{2}}{{r}_{m}^{2}}{B}_{1}$$

(29)

$$\zeta_{2} = p^{2} \varepsilon^{2} - \frac{{2mV_{0} p^{2} a^{2} }}{{\hbar^{2} }} - \frac{{l\left( {l + 1} \right)a^{2} }}{{r_{m}^{2} }}\left( {pB_{1} - B_{2} } \right) - \frac{{0.44V_{0} ma}}{{\hbar^{2} r^{\prime }_{0} }}\left( {j\left( {j + 1} \right) - l\left( {l + 1} \right) - \frac{3}{4}} \right)$$

(30)

Now, Eq. (27) reduces to the NU equation as

$$\frac{{d}^{2}R(z)}{d{z}^{2}}+\frac{(1-pz)}{z(1-pz)}\frac{dR(z)}{dz}+\frac{\left[-{\zeta }_{2}{z}^{2}+{\zeta }_{1}z-{\varepsilon }^{2}\right]}{{\left[z(1-pz)\right]}^{2}}R\left(z\right)=0$$

(31)

By comparing Eq. (31) with the NU equation, we can obtain the energy levels of Eq. (5).

Appendix C

To solve Eq. (6), we consider the following Coulomb term for fluorine (Khordad et al. 2021)

$${V}_{c}\left(r\right)=\frac{8{e}^{2}}{r}$$

(32)

We use the new variable $\phi \left(r\right)=rR(r)$ and obtain

$$\frac{{d}^{2}\phi (r)}{d{r}^{2}}+\frac{2mE}{{\hslash }^{2}}\phi \left(r\right)-\frac{2m}{{\hslash }^{2}}\left[\frac{-{V}_{0}}{1+{e}^{\frac{r-{R}_{0}}{a}}}-0.44{V}_{0}{r}_{0}^{2}\frac{1}{r}\frac{d}{dr}\left(\frac{1}{1+{e}^{\frac{r-{R}_{0}}{a}}}\right)\left(j\left(j+1\right)-l\left(l+1\right)-\frac{3}{4}\right)+\frac{3{e}^{2}}{\pi {\varepsilon }_{0}{R}_{c}}-\frac{{e}^{2}{r}^{2}}{\pi {\varepsilon }_{0}{R}_{c}^{3}}\right]\phi \left(r\right)-\frac{l\left(l+1\right)}{{r}^{2}}=0$$

(33)

We employ another variable as $t={e}^{-\rho r}$ and the following approximation

$$\frac{1}{{r}^{2}}\approx {\rho }^{2}{\left(\frac{{e}^{-\rho r}}{1-{e}^{-\rho r}}\right)}^{2}$$

(34)

We obtain

$${\rho }^{2}{t}^{2}\frac{{d}^{2}R(t)}{d{t}^{2}}+{\rho }^{2}t\frac{dR(t)}{dt}+\left[\frac{2mE}{{\hslash }^{2}}+\frac{2m{V}_{0}p}{{\hslash }^{2}}\left(\frac{t}{1+pt}\right)+\frac{0.44m{V}_{0}p{r}_{0}^{2}{\rho }^{2}}{{\hslash }^{2}}\left(j\left(j+1\right)-l\left(l+1\right)-\frac{3}{4}\right)\frac{{t}^{2}}{(1-t){\left(1+pt\right)}^{2}}-\frac{6m{e}^{2}}{\pi {\varepsilon }_{0}{\hslash }^{2}{R}_{c}}+\frac{2m{e}^{2}}{\pi {\varepsilon }_{0}{\hslash }^{2}{\rho }^{2}{R}_{c}^{3}}\frac{{\left(1-t\right)}^{2}}{{t}^{2}}-l(l+1){\rho }^{2}\frac{{t}^{2}}{{\left(1-t\right)}^{2}}\right]R\left(t\right)=0$$

(35)

We simply above equation and obtain

$$\frac{{d}^{2}R(t)}{d{t}^{2}}+\frac{1}{t}\frac{dR(t)}{dt}+\left[\frac{\frac{2mE}{{\hslash }^{2}{\rho }^{2}}{\left(1-t\right)}^{2}+\frac{0.88m{V}_{0}p}{{\hslash }^{2}{\rho }^{2}}\frac{t{(1-t)}^{2}}{1+pt}}{{t}^{2}{\left(1-t\right)}^{2}}+\frac{\frac{0.44m{V}_{0}p{r}_{0}^{2}}{{\hslash }^{2}}\left(j\left(j+1\right)-l\left(l+1\right)-\frac{3}{4}\right)\frac{{t}^{2}(1-t)}{{\left(1+pt\right)}^{2}}-\frac{6m{e}^{2}}{\pi {\varepsilon }_{0}{\hslash }^{2}{R}_{c}{\rho }^{2}}{\left(1-t\right)}^{2}}{{t}^{2}{\left(1-t\right)}^{2}}+\frac{\frac{2m{e}^{2}}{\pi {\varepsilon }_{0}{\hslash }^{2}{\rho }^{2}{R}_{c}^{3}{\rho }^{4}}\frac{{\left(1-t\right)}^{4}}{{t}^{2}}-l(l+1){t}^{2}}{{t}^{2}{\left(1-t\right)}^{2}}\right]R\left(t\right)=0$$

(36)

To convert the above equation with the NU equation, we use the following approximations (Khordad et al. 2021)

$$\frac{{t\left( {1 - t} \right)^{2} }}{1 + pt} = 0.006595579t^{2} - 0.0132710857t + 0.0067413175$$

(37)

$$\frac{{t^{2} \left( {1 - t} \right)}}{{\left( {1 + pt} \right)^{2} }} = - 3.3579988 \times 10^{ - 9} t^{2} - 0.432019240084 \times 10^{ - 4} t + 0.4341808999 \times 10^{ - 4}$$

(38)

$$\frac{{\left( {1 - t} \right)^{4} }}{{t^{2} }} = 0.981t^{2} - 3.537t + 3.4$$

(39)

By using the above approximation, Eq. (36) convert to the NU equation, and we can obtain the energy levels of Eq. (7).