NEET Chemistry Structure of Atom / परमाणु संरचना Uncertainty principle and Schrodinger wave equation

Uncertainty Principle and Schrodinger Wave Equation

Heisenberg’s uncertainty principle.

(1) One of the important consequences of the dual nature of an electron is the uncertainty principle, developed by Warner Heisenberg.

              (2) According to uncertainty principle “It is impossible to specify at any given moment both the position and momentum (velocity) of an electron”.

              Mathematically it is represented as, \[\Delta x\,.\,\Delta p\ge \frac{h}{4\pi }\]

              Where \[\Delta x=\]uncertainty is position of the particle, \[\Delta p=\]uncertainty in the momentum of the particle

              Now since \[\Delta p=m\,\Delta v\]

              So equation becomes,  \[\Delta x.\,m\Delta v\ge \frac{h}{4\pi }\] or  \[\Delta x\,\times \,\Delta v\ge \frac{h}{4\pi m}\]

              The sign \[\ge \] means that the product of \[\Delta x\]and \[\Delta p\](or of \[\Delta x\]and\[\Delta v\]) can be greater than, or equal to but never smaller than \[\frac{h}{4\pi }.\]If \[\Delta x\]is made small, \[\Delta p\]increases and vice versa.

            (3) In terms of uncertainty in energy, \[\Delta E\]and uncertainty in time \[\Delta t,\]this principle is written as,        \[\Delta E\,.\,\Delta t\ge \frac{h}{4\pi }\]

              Note :q Heisenberg’s uncertainty principle cannot we apply to a stationary electron because its velocity is 0 and position can be measured accurately..

Example: 36 What is the maximum precision with which the momentum of an electron can be known if the uncertainty in the position of electron is \[\pm 0.001{\AA}?\] Will there be any problem in describing the momentum if it has a value of \[\frac{h}{2\pi {{a}_{0}}},\]where \[{{a}_{0}}\]is Bohr’s radius of first orbit, i.e., 0.529Å?

Solution :          \[\Delta x\,.\,\Delta p=\frac{h}{4\pi }\]

                                                   \[\because \] \[\Delta x=0.001{\AA}={{10}^{-13}}m\]

                                      \[\therefore \] \[\Delta p=\frac{6.625\times {{10}^{-34}}}{4\times 3.14\times {{10}^{-13}}}=5.27\times {{10}^{-22}}\]

Example: 37 Calculate the uncertainty in velocity of an electron if the uncertainty in its position is of the order of a 1Å.

Solution : According to Heisenberg’s uncertainty principle

                                                                   \[\Delta v\,.\,\Delta x\approx \frac{h}{4\pi m}\]

                                                                   \[\Delta v\approx \frac{h}{4\pi m.\Delta x}\] \[=\frac{6.625\times {{10}^{-34}}}{4\times \frac{22}{7}\times 9.108\times {{10}^{-31}}\times {{10}^{-10}}}\] \[=5.8\times {{10}^{5}}m\ \,{{\sec }^{-1}}\]

Example: 38 A dust particle having mass equal to \[{{10}^{-11}}g,\]diameter of \[{{10}^{-4}}cm\] and velocity \[{{10}^{-4}}cm\,{{\sec }^{-1}}.\]The error in measurement of velocity is 0.1%. Calculate uncertainty in its positions. Comment on the result.

Solution :                      \[\Delta v=\frac{0.1\times {{10}^{-4}}}{100}=1\times {{10}^{-7}}cm\,{{\sec }^{-1}}\]

                                                                   \[\because \] \[\Delta v\,.\,\Delta x=\frac{h}{4\pi m}\]

                                                                   \[\therefore \] \[\Delta x=\frac{6.625\times {{10}^{-27}}}{4\times 3.14\times {{10}^{-11}}\times 1\times {{10}^{-7}}}=5.27\times {{10}^{-10}}cm\]

                                                                   The uncertainty in position as compared to particle size.

                                                                   \[=\frac{\Delta x}{diameter}=\frac{5.27\times {{10}^{-10}}}{{{10}^{-4}}}=5.27\times {{10}^{-6}}cm\]

                                                                   The factor being small and almost being negligible for microscope particles.

 Schrödinger wave equation.

(1) Schrodinger wave equation is given by Erwin Schrödinger in 1926 and based on dual nature of electron.

(2) In it electron is described as a three dimensional wave in the electric field of a positively charged nucleus.

(3) The probability of finding an electron at any point around the nucleus can be determined by the help of Schrodinger wave equation which is,

\[\frac{\partial {}^{2}\Psi }{\partial x{}^{2}}+\,\frac{\partial {}^{2}\Psi }{\partial y{}^{2}}+\,\frac{\partial {}^{2}\Psi }{\partial z{}^{2}}+\frac{8\pi {}^{2}m}{h{}^{2}}(E-V)\,\Psi =0\]

Where \[x,\,y\] and z are the 3 space co-ordinates, m = mass of electron, h = Planck’s constant,  

                        E = Total energy, V = potential energy of electron, \[\Psi \]= amplitude of wave also called as wave function.

                        \[\partial \] = stands for an infinitesimal change.

(4) The Schrodinger wave equation can also be written as :

\[\nabla {}^{2}\Psi +\frac{8\pi {}^{2}m}{h{}^{2}}(E-V)\,\,\Psi =0\]

   Where \[\nabla \]= laplacian operator.

(5) Physical Significance of \[\Psi \] and \[\Psi {}^{2}\]

(i) The wave function \[\Psi \] represents the amplitude of the electron wave. The amplitude \[\Psi \] is thus a function of space co-ordinates and time i.e. \[\Psi =\Psi (x,\,y,\,\,z......times)\]

(ii) For a single particle, the square of the wave function \[(\Psi {}^{2})\] at any point is proportional to the probability of finding the particle at that point.

 (iii) If \[\Psi {}^{2}\] is maximum than probability of finding \[{{e}^{-}}\] is maximum around nucleus. And the place where probability of finding \[e{}^{-}\] is maximum is called electron density, electron cloud or an atomic orbital. It is different from the Bohr’s orbit.

(iv) The solution of this equation provides a set of number called quantum numbers which describe specific or definite energy state of the electron in atom and information about the shapes and orientations of the most probable distribution of electrons around the nucleus.