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

Ψ=Asin2πxλ\Psi = A \sin \frac{2 \pi x}{\lambda}  u=2πxλu = \frac{2 \pi x}{\lambda} 

dΨdx=dΨdududx=(Acos2πxλ)(2πλ)=A2πλcos2πxλ\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}

d2Ψdx2=d(dΨ)dududx\frac{d^2 \Psi}{dx^2} = \frac{d (d \Psi )}{du} \cdot \frac{du}{dx}  =(A2πλsin2πxλ)(2πλ)==(-\frac{A 2 \pi}{\lambda} \sin \frac{2 \pi x}{\lambda})(\frac{2 \pi}{\lambda})= =π2Aλ2sin2πxλ=- \frac{\pi ^2 A}{\lambda ^2} \sin \frac{2 \pi x}{\lambda}

=yπ2AΨλ2= \frac{-y \pi^2 A \Psi}{\lambda ^2}


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


KE=12mv2=mh22m2λ2=h22mλ2KE = \frac{1}{2} mv^2 = \frac{mh^2}{2m^2 \lambda ^2} = \frac{h^2}{2m \lambda^2}

1λ2=2mK.Eh2\therefore \frac{1}{\lambda ^2} = \frac {2m K.E}{h^2}


d2Ψdx2+8π2mK.EΨh2=0\frac{d^2 \Psi}{dx^2} + \frac{8 \pi ^2 m K.E \Psi}{h^2}=0


Total Energy: KE+PE=EKE + PE = E

KEZe2rKE - \frac{Ze^2}{r} =E=E

KE=E+Ze2rKE = E+ \frac{Ze^2}{r}


d2Ψdx2\frac{d^2 \Psi }{dx^2}+8π2mh2(E+Ze2r)Ψ=0+ \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

2p2p: n=2, l=1n=2, \ l = 1 \rightarrow  Smaller Orbital 

3p3p: n=3, l=1n=3, \ l=1 \rightarrow Larger Orbital

3p*3p has a higher energy than 2p2p because as the nn value incereases the energy increases

Distance \uparrow, Energy \uparrow (E=An2(E= \frac{A}{n^2}))


Quantum Number Questions

Drawing the Orbitals

P - Axes fo through oribtal (3 axes)

d - Axes go between orbital (3 axes)




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