Question

The applications of Schrodinger wave equation are:
(A) It is the bases of wave mechanics
(B) It helps in studying the structure of atom
(C) It shows all the wave like properties of matter
(D) All of the above

Answer 1

Hint:Schrodinger wave equation is the equation of quantum mechanics given by Erwin Schrodinger in 1926. The equation is ${\displaystyle \frac{{\partial }^2\psi }{\partial x^2}}+{\displaystyle \frac{{\partial }^2\psi }{\partial y^2}}+{\displaystyle \frac{{\partial }^2\psi }{\partial z^{\to 2}}}+{\displaystyle \frac{8{\pi }^{2}m(e-v)}{h^2}}\psi =0$

Complete step by step answer: Let us first understand about quantum mechanics and its needs. Classical mechanics based on newton laws of motion was successful in explaining the motion of macroscopic bodies but it failed when applied to microscopic particles. This was mainly because it didn't take into account the concept of dual nature of matter and Heisenberg uncertainty principle. Hence a new branch of science which takes dual nature of matter into consideration has been put forward. This is known as quantum mechanics. Quantum mechanics as developed by Schrodinger in 1926 is based on the wave motion associated with the particles. For wave motion of the electron in three dimensional spaces around the nucleus hence he put forward an equation known after his name as Schrodinger wave equation which is considered as the heart of quantum mechanics. The equation is:-
${\displaystyle \frac{{\partial }^2\psi }{\partial x^2}}+{\displaystyle \frac{{\partial }^2\psi }{\partial y^2}}+{\displaystyle \frac{\partial \psi }{\partial z^2}}+{\displaystyle \frac{8{\pi }^{2}m(e-v)}{h^2}}\psi =0$
Where $\psi$is the amplitude of the wave where coordinates of electrons are (x,y,z), E is the total energy of electrons, V is its potential energy, m is the mass of the electron and h is Planck's constant. ${\partial }^2\psi /d{x}^2$ Represents second derivative of $\psi$ with respect to x and so on.
The wave function $\psi$ for an electron in an atom has no physical significance as such whereas ${\psi }^2$ given the intensity of the electron wave.
Therefore we got to know that option (A) (B) and (C) are the applications of Schrodinger wave equation. Hence option (D) is the correct answer.

Note:Schrodinger wave equations can be shortly written as $\hat{H}\psi =E\psi$ where $\hat{H}$ is a mathematical operator called Hamiltonian operator and whose volume is.
$\hat{H}=-{\displaystyle \frac{h^2}{8{\pi }^{2}m}}\left[{\displaystyle \frac{{\partial }^2}{\partial x}}+{\displaystyle \frac{{\partial }^2}{dy}}+{\displaystyle \frac{{\partial }^2}{d^2}}\right]+V$
The value of E from this equation represents the quantized values of energy which the electrons in the atom can have.