Chapter 1st:- Atomic structure and Periodic table
B.sc 1st year Book
(Page 5)

Schrodinger Wave Equation

The vibrating string behaves like a wave. The motion of the electron is also like a sine wave. Let us consider a sine wave whose wavelength is ‘ λ ‘, ‘ A ‘ is the amplitude at any point along the “x” axis and “y” is its wave function, then

$$ \psi =A\sin \left(\frac{2\pi x}{\lambda }\right)….\left(i\right) $$

Differentiating the above equation with respect to the x, then

$$ \frac{d\psi }{dx}=A\sin \left(\frac{2\pi x}{\lambda }\right)\times \frac{2\pi }{2} $$

$$ \frac{d\psi }{dx}=\frac{2\pi \:}{2}A\sin \left(\frac{2\pi x}{\lambda }\right) $$

Now the second differential coefficient of the above equation is

$$ \frac{d^2\psi }{dx^2}=-\frac{4\pi ^2\:}{\lambda }A\sin \left(\frac{2\pi x}{\lambda }\right),\:\:Since\:\:\left(\psi =A\sin \left(\frac{2\pi \:x}{\lambda \:}\right)\right) $$

$$ \frac{d^2\psi }{dx^2}=-\frac{4\pi ^2\:}{\lambda }\psi \:=\left(ii\right) $$

one dimensional sine wave

This is the equation when the electron will move only in one direction. Now when an electron moves in the three-dimensional space x, y, and z axes, then

$$ \frac{d^2\psi \:}{dx^2}\:+\frac{d^2\psi \:^2\:}{dy^2}+\frac{d^2\psi \:\:}{dx^2}=-\frac{4\pi \:^2}{2^2}v $$ ……….. (iii)

Now

$$ Now\:\frac{d^2\:}{dx^2}\:+\frac{d^2\:}{dy^2}+\frac{d^2\:\:}{dx^2}\:is\:shortened\:to\:\Delta ^2 $$

Where x, y and z are three cartesian axes;

$$ \:\frac{d^2y\:}{dx^2}\: $$ is rate of the change of wave location y with x;

$$ \frac{d^2\psi }{dx^2}\:is\:rate\:of\:the\:change\:of\:\frac{dv}{dx}\:with\:x. $$

Similarly

$$ \frac{d\psi }{dy}\:and\:\frac{d\psi \:}{dz}\:are\:the\:rate\:of\:changes\:of\:y\:with\:y\:and\:z\:respectively $$

$$ According\:to\:de-broglie\:equation\:we\:know\:that\:\lambda =\frac{h}{mv} $$

Or

$$ \lambda ^2=\frac{h^2}{m^2v^2} $$

Where m is mass and v is the velocity of moving electrons

Hence,

$$ \Delta ^2y=-\frac{4\pi ^2m^2y^2}{h^2} $$

or

$$ \Delta ^2y=-\frac{4\pi ^2m^2y^2}{h^2}\psi =0 $$

The total energy of a moving electron is the sum of kinetic and potential energies, hence

(‘V’ is Potential energy)

Putting the value of v2 in equation (iv) we got

$$ E=\frac{1}{2}mv^2\:+\:V $$

$$ v^2=\frac{2\left(E-V\right)}{m} $$

$$ \Delta ^2\psi +\frac{8\pi ^2m}{h^2}\left(E-V\right)\psi =0 $$

This is a well-known form of the Schrodinger wave equation
$$ Multiplying\:equation\:\left(v\right)\:by\:a\:quantity\:\frac{h^2}{8\pi ^2m},\:we\:get $$
$$ \frac{h^2}{8\pi ^2m}\Delta ^2\psi +E\psi -Vy=0 $$
Or
$$ \left(V-\frac{h^2}{8\pi ^2m}\Delta ^2\right)\psi =E\psi $$
Or
Hy = Eψ
Where ‘H’ is the Hamiltonian operator and ‘E’ is the Eigenvalue (operator form of Schrodinger wave equation).

Significance of wave function

The physically possible solution of the Schrodinger wave equation must have certain properties of the wave function.
1- ψ the mist is continuous and finite.
2- ψ must be single-valued.
3. The probability of finding the electron in overall space from +∞ and -∞ must be equal to one. Ψ2 is the probability of finding the electron at a point.
$$ \int _{-\infty }^{+\infty }\:y^2dxdydz=1 $$
There are several waves functions ψ1, ψ2, ψ3 which will satisfy these conditions to the wave equation. Each of these wave functions has corresponding energy E1, E2, E3 respectively. Thus. etc is called “an orbital’.
For a particular atom, there are a number of solutions to the Schrodinger wave equation that are sure acceptable and each orbital is designated by ‘n’, ‘l’, and ‘m’ which are called principal, azimuthal, and magnetic quantum numbers respectively.
In the first group of solutions, the probability of finding the electron sai symboldepends on the radial distance ‘ r ‘ and is the same in all directions l.e.  (spherical orbit).
The second group of solutions depends upon ‘ ‘r‘ as well as on the direction in space along the x,y, and z axes.
ux=I(r)f(x)
uy=I(r)f(y)
ψz=I(r)f(z)
There are three p-orbitals in the degenerate state obtained for n = 2,4,5….. Whereas the third group of solutions to the wave equation show this group of orbitals has I =2  and is known as d-orbitals for each of the values of n = 3,4,5… Lastly, a set of solutions occur l = 3 and n = 4,5,6….  which are called f-orbitals.

Radial and Angular Functions ;

In the case of hydrogen being hydrogen-like atoms such as (He+, Li2+, Be3+,..) which have only The cartesian coordinate and are comverted into polar coordinate with the help of and . one electron; the Schrodinger wave equation can be solved completely in terms of the conversion of cartesian coordinates into polar coordinates. The cartesian coordinates x, y, and z are converted into polar coordinates with the help x, y and z.

The Schrodinger wave equation.

$$ \Delta ^2y+\frac{8\pi ^2m}{h^2}\left(E-V\right)\psi =0 $$

Where

$$ \Delta ^2\psi =\left(\frac{d^2}{dx^2}+\frac{d^2}{dy^2}+\frac{d^2}{dz^2}\right) $$

$$ \frac{1d}{r^2dr}\left(r^2\frac{d\psi \:}{dx}\right)+\frac{1}{r^2sin^2\theta \:}\times \frac{d^2\psi \:^2}{d\phi \:^2}+\frac{1}{r^2sin\theta \frac{d}{d\theta }}\left(sin\theta \:\frac{d\psi \:}{d\theta \:}\right)+ $$
$$ \frac{8\pi ^2m}{h^2}\left(E+\frac{e^2}{r}\right)\psi r=0 $$
$$ Since\:\left(V=-\frac{Ze^2}{r}\:and\:z=1\:for\:H\right) $$
The Solution of the this equation is in the form of ψ=R(r)Θ(θ)∅(∅)
Where R(r) is a function that depends on the quantum numbers n and is a function that depends on the quantum numbers I and is a function that depends on the quantum number m only.
R(r) is a function of the radial part of the wave function where asand are called the ‘angular parts of the wave function’, ‘ r ‘ is the radial distance of point , the position of an electron from the nucleus.
the position of electron from nucleus.

Radial Probability distribution Curve :

The radial function F has no physical meaning but is the probability of finding the electron in a small volume ‘ dV ‘ near the point at which R is measured. For a given value of ‘ r ‘, the value of The probability of an electron at a distance and is known as the radial distribution function. When ‘ r ‘ is plotted against for hydrogen atom, we get the following curve-
 Redial probability density for some orbitais of the hydrogen atom. Ordinate is proportional to and all distributions are to the same scale.
 The redial probability density for some orbitals of the hydrogen atom. The ordinate is proportional to and all distributions are to the same scale.
  1. The probability of finding the electron when r=0 ( 1.e. near the nucleus) is zero,
  2. The greater the value of the principal quantum number ‘n’, the probability of finding the electron is farther from the nucleus.
  3. The probability of finding the electron for 18 orbital is at a distance of 0.529 A from the nucleus and is in good agreement with Bohr’s first circular orbit for the hydrogen atom. The probable distance for 2s or 2p orbital is 2,1 A.

Angular Probability distribution curve :

The angular function and depend only on the direction in the three-dimensional space and is independent of the distance from the nucleus. The probability of finding the electron at a distance ‘r’ from the nucleus and in a given directionThe probablity of finding the electron at a distance 'r' from the nucleus and in a given cirection.

The probablity of finding the electron at a distance 'r' from the nucleus and in a given direction

Thus, is the probability of finding the electron in a given direction, at any distance ‘ r ‘ from the nucleus to infinity. The angular part of the wave function arises (due to drawing on polar coordinates) and gives the shape of orbitals. They show a positive or negative sign relating to the symmetry of the angular function. For bonding, the overlap of the orbital of a similar sign must take place.

1: It is easier to visualize the boundary surface of orbitals in the form of a solid shape having 90% of the electron density, For p – orbitals the electron density at the nucleus is zero. According to some texts, p-orbitals are two spheres that do not touch each other as shown in figure 1.09.

2: sorbitals are spherically symmetrical and most of the electron density lie at the surface boundary of sphere. 2 s and 3 s also have a symmetrical sphere of increasing size. The drawing of 2p, 3p, 4p, 3d, 4d etc. also show the symmetry as shown in figure 1.09.

3: The probability of finding the electron towards direction is  and . The diagram is figure are the angular part of the wave function.

4: d-orbitals are five in number in which dxy, dyz, dzx, and orbitals have the same shape but differ in their orientationorbitals have the same shape but differ in their orientation. dxy, dyz, dzx lobes lie towards, yz and zx planes respectively,orbitals have the same shape but differ in their orientation lobes lying along and axes and lobes lie along the axis but the electron density also lies in the plane.

Angular part of the wave function A(0, ) for the 1s,2p and 3d orbitals for a hydrogen atom shown as polar diagrams

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