# Lecture 9: Derivation of Schrodinger Equation and Probability of Finding an Electron

﻿$*$﻿ Know how to explain what happens when an atoms becomes an ion

Size Depends On:

• Position on periodic table
• ﻿$+/-$﻿ ion

### Representing an atom by a mathematical equation: Schrodinger

#### Derivation of Schrodinger Equation

﻿$\Psi = A \sin \frac{2 \pi x}{\lambda}$﻿ ﻿$u = \frac{2 \pi x}{\lambda}$﻿

﻿$\frac{d \Psi}{dx} = \frac{d \Psi}{du} \cdot \frac{du}{dx} = (A \cos \frac{2 \pi x}{\lambda} )(\frac{2 \pi}{\lambda} )= \frac{A 2 \pi}{\lambda} \cos \frac{2 \pi x}{\lambda}$﻿

﻿$\frac{d^2 \Psi}{dx^2} = \frac{d (d \Psi )}{du} \cdot \frac{du}{dx}$﻿ ﻿$=(-\frac{A 2 \pi}{\lambda} \sin \frac{2 \pi x}{\lambda})(\frac{2 \pi}{\lambda})=$﻿ ﻿$=- \frac{\pi ^2 A}{\lambda ^2} \sin \frac{2 \pi x}{\lambda}$﻿

﻿$= \frac{-y \pi^2 A \Psi}{\lambda ^2}$﻿

de Brogile: ﻿$mv = \frac{n}{\lambda}$﻿ ﻿$v = \frac{h}{m \lambda}$﻿

﻿$KE = \frac{1}{2} mv^2 = \frac{mh^2}{2m^2 \lambda ^2} = \frac{h^2}{2m \lambda^2}$﻿

﻿$\therefore \frac{1}{\lambda ^2} = \frac {2m K.E}{h^2}$﻿

﻿$\frac{d^2 \Psi}{dx^2} + \frac{8 \pi ^2 m K.E \Psi}{h^2}=0$﻿

Total Energy: ﻿$KE + PE = E$﻿

﻿$KE - \frac{Ze^2}{r}$﻿ ﻿$=E$﻿

﻿$KE = E+ \frac{Ze^2}{r}$﻿

﻿$\frac{d^2 \Psi }{dx^2}$﻿﻿$+ \frac {8 \pi^2 m}{h^2} (E + \frac{Ze^2}{r}) \Psi = 0$﻿

#### Probability of finding an electron:

90% of electrons are found in the light green area

• When the electrons are dense ﻿$\rightarrow$﻿ less electrons
• When the electrons are less dense and more spread out there are more electrons

Graph looks like:

Not like Bohr says: 90% not 100%

#### What is the difference between 2p and 3p

﻿$2p$﻿: ﻿$n=2, \ l = 1$﻿ ﻿$\rightarrow$﻿ Smaller Orbital

﻿$3p$﻿: ﻿$n=3, \ l=1$﻿ ﻿$\rightarrow$﻿ Larger Orbital

﻿$*3p$﻿ has a higher energy than ﻿$2p$﻿ because as the ﻿$n$﻿ value incereases the energy increases

Distance ﻿$\uparrow$﻿, Energy ﻿$\uparrow$﻿ ﻿$(E= \frac{A}{n^2}$﻿﻿$)$﻿

### Drawing the Orbitals

P - Axes fo through oribtal (3 axes)

d - Axes go between orbital (3 axes)