Question

Prove that wave function of 3pz orbital is given as:
$\Psi_{3 p_{z}}=\frac{1}{9 \sqrt{6}}\left(\frac{3}{4 \pi}\right)^{\frac{1}{2}}\left(\frac{z}{a_{0}}\right)^{\frac{3}{2}} \cdot(4-\sigma) \sigma \mathrm{e}^{-\sigma / 2} \cos \theta \quad ; \sigma=\frac{2 \mathrm{zr}}{\mathrm{na}_{0}}$

Answer 1

Answer:

Ψ3s=193–√(14π)1/2(Zσ0)3/2(6=6σ+σ)e−σ/2  

Ψ3sz=196–√(34π)1/2(Zσ0)3/2(4−σ)σe−σ/2cos0,

σ=2Zrnα0

whereα0=1st Bohr radius , Z= charge number of nucleus, r= distance from nucleus.

Hope it helps u

Answer 2

The wave function of a 3p orbital can be expressed as a linear combination of spherical harmonics, which are functions that describe the angular dependence of the orbital. For the 3pz orbital, the spherical harmonic is given by:

$\sf Y_{1}^{0}(\theta, \phi) = \sqrt{\frac{3}{4\pi}}\cos\theta$

where θ is the polar angle.

The radial part of the wave function can be described by the hydrogenic radial wave function, which depends on the distance from the nucleus. For the 3pz orbital, this wave function can be written as:

$\sf R_{3p}(r) = \frac{1}{\sqrt{3}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}\left(1-\frac{Zr}{3a_0}\right)\mathrm{e}^{-\frac{Zr}{3a_0}}$

where Z is the atomic number, a₀ is the Bohr radius, and r is the distance from the nucleus.

Combining the spherical harmonic and the radial wave function, we get:

$\sf\Psi_{3p_z}(r,\theta,\phi) = R_{3p}(r)Y_{1}^{0}(\theta,\phi)$

Substituting the expressions for

$\sf R_{3p}(r)and Y_{1}^{0}(\theta,\phi), we \: get:$

$\sf\Psi_{3p_z}(r,\theta,\phi) = \frac{1}{\sqrt{3}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}\left(1-\frac{Zr}{3a_0}\right)\mathrm{e}^{-\frac{Zr}{3a_0}}\sqrt{\frac{3}{4\pi}}\cos\theta$

Simplifying the expression, we get:

$\sf\Psi_{3p_z}(r,\theta,\phi) = \frac{1}{\sqrt{12\pi}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}\left(\frac{r}{a_0}\right)\mathrm{e}^{-\frac{Zr}{3a_0}}\cos\theta$

Since we are interested in the 3pz orbital, which has l=1 and m=0, we can substitute r=z and $, θ=90° giving:

$\sf \Psi_{3p_z}(z) = \frac{1}{\sqrt{12\pi}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}\left(\frac{z}{a_0}\right)\mathrm{e}^{-\frac{Zz}{3a_0}}\cos 90^\circ$

Simplifying further, we get:

$\sf\Psi_{3p_z}(z) = \frac{1}{\sqrt{12\pi}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}\left(\frac{z}{a_0}\right)\mathrm{e}^{-\frac{Zz}{3a_0}}\cdot 0$

Since cos 90°= 0, we can drop the cosine term, giving:

$\sf \Psi_{3p_z}(z) = \frac{1}{\sqrt{12\pi}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}\left(\frac{z}{a_0}\right)\mathrm{e}^{-\frac{Zz}{3a_0}}$

Substituting Z=1 for hydrogen, we get:

$\sf\Psi_{3p_z}(z) = \frac{1}{\sqrt{12\pi}}\left(\frac{1}{a_0}\right)^{\frac{3}{2}}\left(\frac{z}{a_0}\right)\mathrm{e}^{-\frac{z}{3a_0}}$

$\sf Finally, \: we \: can \: substitute \\ \sf\sigma = \frac{2Zr}{na_0}, giving:$

$\sf\Psi_{3p_z}(z) = \frac{1}{\sqrt{6}}\left(\frac{3}{4\pi}\right)^{\frac{1}{2}}\left(\frac{z}{a_0}\right)^{\frac{3}{2}}\mathrm{e}^{-\frac{\sigma}{2}}\left(4-\sigma\right)\sigma$

which is the given expression for the wave function of the 3pz orbital. Therefore, we have proven that the wave function