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
$$ \frac{d^2\psi }{dx^2}=-\frac{4\pi ^2\:}{\lambda }A\sin \left(\frac{2\pi x}{\lambda }\right),\:\:Since\:\:\left(\psi =A\sin \left(\frac{2\pi \:x}{\lambda \:}\right)\right) $$
$$ \frac{d^2\psi }{dx^2}=-\frac{4\pi ^2\:}{\lambda }\psi \:=\left(ii\right) $$
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 v^{2} in equation (iv) we got
$$ E=\frac{1}{2}mv^2\:+\:V $$
$$ v^2=\frac{2\left(E-V\right)}{m} $$
$$ \Delta ^2\psi +\frac{8\pi ^2m}{h^2}\left(E-V\right)\psi =0 $$
Significance of wave function :
2- ψ must be single-valued.
3. The probability of finding the electron in overall space from +∞ and -∞ must be equal to one. Ψ^{2} is the probability of finding the electron at a point.
In the first group of solutions, the probability of finding the electron depends 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.
Radial and Angular Functions ;
Related Topic | Atomic Structure and Periodic Table |
The Schrodinger wave equation.
$$ \Delta ^2y+\frac{8\pi ^2m}{h^2}\left(E-V\right)\psi =0 $$
Where
$$ \Delta ^2\psi =\left(\frac{d^2}{dx^2}+\frac{d^2}{dy^2}+\frac{d^2}{dz^2}\right) $$
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.
Radial Probability distribution Curve :
The redial probability density for some orbitals of the hydrogen atom. The ordinate is proportional to and all distributions are to the same scale.
- The probability of finding the electron when r=0 ( 1.e. near the nucleus) is zero,
- The greater the value of the principal quantum number ‘n’, the probability of finding the electron is farther from the nucleus.
- 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 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: s–orbitals 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 orientation. dxy, dyz, dzx lobes lie towards, yz and zx planes respectively, lobes lying along and axes and lobes lie along the axis but the electron density also lies in the plane.