(a) Using Equation 4.88, work out the first four Laguerre polynomials.

(b) Using Equations 4.86, 4.87, and 4.88, find $\mathit{v}\left(\rho \right)$, for the case $\mathit{n}\mathbf{=}\mathbf{5}\mathbf{,}\mathit{I}\mathbf{=}\mathbf{2}$.

(c) Find $\mathit{v}\left(\rho \right)$again (for the case role="math" localid="1658315521558" $\mathbf{n}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{I}\mathbf{=}\mathbf{2}$), but this time get it from the recursion formula (Equation 4.76).

$$\begin{gathered} L_q\left(x\right)=\frac{e^q}{q!}\left(\frac{d}{dx}{)}^q\right(e^{-x}-x^9)(4.88\left) \\ v\right(\rho )={L_{n-}^{2l+1}}_{l-1}(4.86\left) \\ {L}_q^p\right(x)\equiv (-1{)}^p{\left(\frac{d}{dx}\right)}^{\rho }{L}_{p+q}\left(x\right)(4.87) \\ {c}_{j+1}=\frac{2(j+l+1-n)}{(j+1)(j+2l+2)}{c}_j(4.76) \end{gathered}$$

The probability of finding an electron in some finite volume element around a point at a distance of r from the nucleus is given by the radial wave function R(r), which is simply the value of the wave function at some radius r.

(a)

The associated Laguerre polynomial is given by equation as

For $q=0$

$$\begin{gathered} L_0=e_x{e}_{-x} \\ {L}_0=1 \end{gathered}$$

For$q=1$

$$\begin{gathered} L_1=e_{xdxd}\left(e_{-x}x\right) \\ =e_x\left[e_{-x}\right)-e_{-x}x] \\ =1-x. \end{gathered}$$

for $q=2$

$$\begin{gathered} L_2=\frac{e^x}{2}{\left(\frac{d}{dx}\right)}^2\left(e^x{x}^2\right) \\ =\frac{e^x}{2}\frac{d}{dx}\left(2x{e}^{-x}-e^{-}{x}^2\right) \\ =\frac{e^x}{2}\left(2e^{-x}-2x{e}^{-x}+e^{-x}{x}^2-2x{e}^{-x}\right) \\ =1-2x+\frac{1}{2}{x}^2 \end{gathered}$$

for $q=3$:

$$\begin{gathered} L_3=\frac{e^x}{6}{\left(\frac{d}{dx}\right)}^3\left(e^{-x}{x}^3\right) \\ =\frac{e^x}{6}{\left(\frac{d}{dx}\right)}^2\left(-e^{-x}{x}^3+3x^2{e}^{-x}\right) \\ =\frac{e^x}{6}\frac{d}{dx}\left(e^{-x}{x}^3+3x^2{e}^{-x}3x^2{e}^{-x}+6x{e}^{-x}\right) \\ =\frac{e^x}{6}\left(-e^{-x}{x}^3+3x^2{e}^{-x}+6x^2{e}^{-x}-12x{e}^{-x}-6x{e}^{-x}+6e^{-x}\right) \\ =1-3x+\frac{3}{2}{x}^2-\frac{1}{6}{x}^3 \end{gathered}$$

Thus,

$\left\{\begin{array}{cccc}{L}_0& =1& & \\ L_1& =1-x& & \\ L_2& =2-4x+x^2& & \\ L_3& =6-18x+9x^2& -x^3& \end{array}\right.\begin{array}{}\end{array}$

(b)

Need to write $v\left(\rho \right)$using,

$$\begin{gathered} v\left(\rho \right)=L_{25}\left(2\rho \right) \\ {L}_{25}\left(x\right)=(-1{)}_{5\left(dxd\right)5}{L}_7(x) \end{gathered}$$

Need to find $L_7\left(x\right)$as:

$$\begin{gathered} L_7\left(x\right)=e^x{\left(\frac{d}{dx}\right)}^7\left(x^7{e}^{-x}\right)=\frac{e^x}{7}!{\left(\frac{d}{dx}\right)}^6\left(7x^6{e}^{-x}-x^7{e}^{-x}\right) \\ =e^x{\left(\frac{d}{dx}\right)}^5\left(42x^5{e}^{-x}-7x^6{e}^{-x}+x^7{e}^{-x}\right) \\ =e^x{\left(\frac{d}{dx}\right)}^4\left(210x^4{e}^{-x}-42x^5{e}^{-x}-84x^5{e}^{-x}+14x^6{e}^{-x}+7x^6{e}^{-x}-x^7{e}^{-x}\right) \\ =e^x{\left(\frac{d}{dx}\right)}^2\left(2520x^2{e}^{-x}-\left(840+3360\right)x^3{e}^{-x}+\left(840+1260\right)x^4{3}^{-x}-\left(252+168\right)x^5{e}^{-x}+\left(228+7\right)x^6{e}^{-x}-x^7{e}^{-x}\right) \end{gathered}$$

$$\begin{gathered} =e^x{\left(\frac{d}{dx}\right)}^2\hspace{0.17em}(2520x^2{e}^{-x}-\left(840+3360\right)x^3{e}^{-x} \\ =\left(840+1260\right)x^4{e}^{-x}-\left(252+168\right)x^5{e}^{-x}+\left(28+7\right)x^6{e}^{-x}-x^7{e}^{-x}) \\ {L}_7\left(x\right)=e^x[5040e^{-x}-\left(5040+30240\right)e{x}^{-x}+\left(15120+37800\right)x^2{e}^{-x} \\ {L}_7\left(x\right)=-\left(12600+8400+8400\right)x^3{e}^{-x}+\left(2100+2100+3150\right)x^4{e}^{-x}-\left(630+252\right)x^5{e}^{-x}+\left(42+7\right)x^6{e}^{-x}-x^7{e}^{-x}] \\ =\left(5040-25280x+52920x^2-29400x^3+7350x^{4}882x^5+49x^6-x^7\right) \end{gathered}$$

Now, to take the 5^th derivative to this to find $L_2^5$any variable x with power less than 5 will vanish, therefore

localid="1658377935803" $$\begin{gathered} L_2^5=-{\left(\frac{d}{dx}\right)}^5\left(-882x^5+49x^6-x^7\right) \\ =-\left[-882\left(5.4.3.2\right)+\left(49\times 12\right)x+42x^2\right] \\ =60\left[\left(882\times 2\right)-\left(49\times 12\right)x+42x^2\right] \\ =2520\left(42-14x+x^2\right) \end{gathered}$$

Thus,

$$\begin{gathered} v\left(\rho \right)=2520\left(42-28\rho +4{\rho }^2\right) \\ =5040\left(21-14\rho +2{\rho }^2\right) \end{gathered}$$

(c)

In this part we will repeat part (b) but using the equations:

$$\begin{gathered} v\left(\rho \right)={\sum }_{j=0}^{\infty }{c}_j{p}^j \\ {c}_{j+1}=\frac{2(j+l+1-n)}{(j+1)(j+2l+2)}{c}_j(4.76) \end{gathered}$$

For $n=5$and$I=2$

$v_{52}\left(\rho \right)=c_0+c_1\rho +c_2{\rho }^2$

Where the constants can be determined using the above equation as:

$c_1=\frac{2(3-5)}{\left(1\right)\left(6\right)}{c}_0=-\frac{2}{3}{c}_0.$

$$\begin{gathered} c_2=\frac{2(4-5)}{\left(2\right)\left(7\right)}{c}_1=-\frac{1}{7}{c}_1=\frac{2}{21}{c}_0 \\ {c}_3=\frac{2(5-5)}{\left(3\right)\left(8\right)} \end{gathered}$$

Substitute into, we get

$$\begin{gathered} v\left(\rho \right)=c_0-\frac{2}{3}{c}_0\rho +\frac{2}{21}{c}_0{\rho }^2 \\ =\frac{c_0}{21}(21-14\rho +2{\rho }^2) \end{gathered}$$