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Atomic Structure

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Overview

Thomson's Experiment

Bohr's Model

Schrödinger's Equation

Overview

The structure of the modern atom is explored through three key advances:
   
(1) J. J. Thomson's third cathode-ray experiment (1897)
(2) Niels Bohr's model for one-electron atoms (1913)
(3) Erwin Schrödinger's equation for quantum atoms and molecules (1925)
   
Thomson, Bohr, and Schrödinger would receive Nobel Prizes for their work in 1906, 1922, and 1933 respectively.

 

Thomson's Experiment

In 1897 J. J. Thomson in his third cathode-ray-tube experiment used electric and magnetic fields to measure the charge-to-mass ratio of the electron.  

Figure 1: Schematic of a cathode ray tube with deflection from an applied electric field (Author: Kurzon (Public Domain))


The applied voltage causes a stream (cathode ray) of negatively charged particles (electrons) to flow through the tube.  If the forces from the electric and magnetic fields are chosen to balance, preventing the deflection of the cathode ray, then the charge-to-mass ratio of the particles can be computed from the formula

qm =E22 VB2

Because the kinetic energy of the electron must be equal to the charge of the electron multiplied by the applied voltage, we have

12mv2=qV

or after solving for the velocity

v = 2qVm

Using this formula, we can compute the velocity of the electrons for a given applied voltage.

 

We load the Units package with Maple's with command

withUnitsSimple:

 

We also load the ScientificConstants package and define the following scientific constants that we will need in this part of the lesson where the subscript SI indicates that we are using Standard International (SI) units: electron charge (e), electron mass (m), and the speed of light (c).  The values of these constants are built into the Maple engine.

withScientificConstants:eSI  evalfConstant'e',units;mSI  evalfConstant'me',units;cSI  evalfConstant'c',units;

eSI1.60217662010−19C

mSI9.10938356010−31kg

cSI2.99792458108ms

(2.1)

 

(a) What is the velocity when a voltage of 2 V (Volts) is applied to the cathode ray tube?  
   
Answer: We can substitute a voltage of 2 V as well as the charge and the mass into the expression for the velocity  

vSI  subsm=mSI,q=eSI,V=2 Unit'volt', sqrt2.0qVm;

vSI838765.7651CVkg

(2.2)

 

Simplifying the units, we have

vSI  combinevSI, 'units';

vSI838765.7651ms

(2.3)

 

(b) What is the fraction of this velocity to the speed of light?

 

Answer: We can compute the ratio of the velocity to the speed of light constant

vSIcSI;

0.002797821435

(2.4)

 

The velocity of the electron is only 3/1000 of the speed of light. (Note that we are not considering relativistic effects).

 

(c) Changing the numbers in (a), determine the velocity when a voltage of 120 V (Volts) is applied to the cathode ray tube.

 

(d) Changing the numbers in (b), what is the fraction of the velocity in (c) to the speed of light?

 

Bohr's Model

In 1913 Niels Bohr proposed a model for the hydrogen atom in which electrons rotate around the positively charge nucleus in discrete circular orbits.  Each orbit has an energy with transitions between orbits corresponding to the emission and absorption of radiation.  The predicted frequencies of emission matched the measured emission spectra of hydrogen gases.  The orbits are labeled by positive integers n known as the principal quantum numbers.

Figure 2: Sketch of the Bohr model's circular orbits with emission of radiation (Author: JabberWok (CC BY-SA 3.0 no changes))

 

The energy of each orbit can be computed from the following formula

En=Z2n2R__H

where the positive integer n is the quantum number, Z is the number of protons, and R__H is the Rydberg constant.
   
The radius of each orbit can be computed from the following formula

r = a0n2Z

where  a__0 is a constant known as the Bohr radius

a__0__ = 5.291772101011m

 

We define additional scientific constants that we will need in this parts of the lesson: Planck's constant (h), Rydberg's constant (RH), and the Bohr radius (a0).

hSI  evalfConstant'h',units;RSI  evalfConstant'h',unitsConstant'c',unitsConstant'Rinfinity',units;a0SI  evalfConstant'a0',units;

hSI6.62607004010−34m2kgs

RSI2.17987232510−18J

a0SI5.29177210510−11m

(3.1)

 

We can define the energy as a function of n, Z, and R__H

energy  n,Z,R  Z2Rn2;

energyn,Z,RZ2R`/`n2

(3.2)

 

(a) What are the first four energies of He+ (note that Z = 2)?

 

Answer: We can easily generate the first four energies using a for-loop in Maple

for i to 4 do   Ei  energyi,2,RSI; end do;

E18.71948930010−18J

E22.17987232510−18J

E39.68832144410−19J

E45.44968081210−19J

(3.3)

 

(b) How much energy is emitted when an electron changes from the n=4 orbit to the n=1 orbit of He+?

 

Answer: We can easily compute the energy difference

E1E4;

8.17452121910−18J

(3.4)

 

(c) What is the frequency of the emitted radiation in (b)?

 

Answer: We know from Einstein (1906) that

E = hv

Therefore,

nu  absE1E4hSI;

ν1.23369073510161s

(3.5)

 

(d) In what region of the electromagnetic spectrum does this frequency lie (i.e. visible or ultraviolet)?

 

We can define the radius as a function of n and Z

radius  n,Z  a0SIn2Z;

radiusn,Za0SIn2`/`Z

(3.6)

 

(e) Changing the calculations in (b), determine how much energy is emitted when an electron changes from the n=4 orbit to the n=3 orbit of He+.

 

(f) What is the frequency of the emitted radiation in (e)?

 

(g) In what region of the electromagnetic spectrum does this frequency lie (i.e. visible or ultraviolet)?

 

(h) What are the radii of the first four orbits of He+?

 

Answer: We can easily generate the first four radii using a for-loop in Maple

for i to 4 do   ri  radiusi,2; end do;

r12.64588605210−11m

r21.05835442110−10m

r32.38129744710−10m

r44.23341768410−10m

(3.7)

 

(i) Draw a sketch of these first four orbits.

 

Answer: We can sketch the orbits using Maple's plot facilities. 

for i to 4 do   pi  plottoolscircleop1, ri, color=red: end do:plotsdisplayp1,p2,p3,p4; 

 

(j) Are the orbits getting closer together or further apart with increasing quantum number n?

 

Schrödinger's Equation

Erwin Schrödinger in 1925 derived a differential equation whose solution for an atom or molecule yields energies as well as electron probability distributions.  Although the energies from the Bohr model are correct for one-electron atoms, the circular orbits of the electrons are not correct.  In Schrödinger's equation, Bohr's notion of orbit is replaced by a probability distribution of electrons, known as an orbital.  The probability of finding an electron at a given position in the orbital, known as the orbital's electron density, is proportional to the square of the orbital evaluated at that position.  Using the Maple Quantum Chemistry Toolbox, we can solve the Schrödinger equation for atoms and molecules.  In this section we examine the orbital densities resulting from the solution of the two-electron helium atom.    

 

First, we load the QuantumChemistry package using Maple's with command

withQuantumChemistry;

AOLabels,ActiveSpaceCI,ActiveSpaceSCF,AtomicData,BondAngles,BondDistances,Charges,ChargesPlot,CorrelationEnergy,CoupledCluster,DensityFunctional,DensityPlot3D,Dipole,DipolePlot,Energy,ExcitationEnergies,ExcitationSpectra,ExcitationSpectraPlot,ExcitedStateEnergies,ExcitedStateSpins,FullCI,GeometryOptimization,HartreeFock,Interactive,Isotopes,MOCoefficients,MODiagram,MOEnergies,MOIntegrals,MOOccupations,MOOccupationsPlot,MOSymmetries,MP2,MolecularData,MolecularGeometry,NuclearEnergy,NuclearGradient,OscillatorStrengths,Parametric2RDM,PlotMolecule,Populations,RDM1,RDM2,RTM1,ReadXYZ,Restore,Save,SaveXYZ,SearchBasisSets,SearchFunctionals,SkeletalStructure,Thermodynamics,TransitionDipolePlot,TransitionDipoles,TransitionOrbitalPlot,TransitionOrbitals,Variational2RDM,VibrationalModeAnimation,VibrationalModes,Video

(4.1)

 

Second, we define the coordinates of the helium atom, placing it at the origin

He He,0,0,0; 

HeHe,0,0,0

(4.2)

 

Third, we use the command HartreeFock to solve the Schrödinger equation for He

data   HartreeFockHe, basis=cc-pVTZ;

table%id=18446744617867454654

(4.3)

 

Fourth, we visualize the first orbital from the solution, known as the 1s orbital.   With the command DensityPlot3D we can plot its electron density in 3D; the command can take several seconds to return as it generates an accurate electron probability distribution

DensityPlot3DHe, data, basis=cc-pVTZ, orbitalindex=1; 

Note that in the 1s orbital the electron is most likely found in a sphere about the nucleus.  The electron is not confined to a shell as in the Bohr model!

 

We can visualize the electron density of the second orbital, known as the 2s orbital, by changing the keyword orbitalindex to 2

DensityPlot3DHe, data, basis=cc-pVTZ, orbitalindex=2;

 

Similarly, we can visualize the third, fourth, and fifth orbitals, known as the 2px, 2py, and 2pz orbitals, which are degenerate in energy with the 2s; all four orbitals, 2s, 2px, 2py, and 2pz, roughly correspond to the n=2 orbit in the Bohr model.  The 2px, 2py, and 2pz orbitals have a probability density of zero at the nucleus.  You can click on the orbital plots and rotate them!

DensityPlot3DHe, data, basis=cc-pVTZ, orbitalindex=3; 

DensityPlot3DHe, data, basis=cc-pVTZ, orbitalindex=4;

DensityPlot3DHe, data, basis=cc-pVTZ, orbitalindex=5;

 

Finally, the 6-10 orbitals, known as 3d orbitals, have the same energy and roughly correspond to the n=3 orbit in the Bohr model.  The electron density of the 7th orbital is given by

DensityPlot3DHe, data, basis=cc-pVTZ, orbitalindex=7;

 

In this case the probability of finding the electron forms an image like 4-leaf clover!

 

(a) Change the orbitalindex to 8 and plot (and then sketch in your assignment) the resulting 3d orbital density. (Hint: You may need to rotate the plot to obtain the best picture!)

 

(b) Change the orbitalindex to 9 and plot (and then sketch in your assignment) the resulting 3d orbital density. (Hint: You may need to rotate the plot to obtain the best picture!)