Quantum theory and models of the atomeggnchips.bucket.s3-website-eu-west-1.amazonaws.com/.../Photoelectric2.pdf · Quantum theory and models of the atom I Radiation intensity is proportional

Quantum theory and models of the atom

Guess now.It has been found experimentally that:

(a) light behaves as a wave;

(b) light behaves as a particle;

(c) electrons behave as particles;

(d) electrons behave as waves;

(e) all of the above are true;

(f) none of the above are true.


I Radiation intensity is proportional to T 4

I Materials glow white/yellow at 2000K.


I The Wavelengths of the peaks follow λpT = 2.90× 10−3

I This is Wien’s law.

I Draw a graph of T against λ.

I What type of curve is it?


I Example: Sun temperature

I If λp =500 nm for the Sun, what is the temperature

T =2.90x10−3

500x10−9= 6000K



I Example: Star colour

I The surface temperature of a star is 32,500K. What colourwould the star appear?

I Answer:

λp =2.x10−3

32, 500= 89.2nm

I This is the UV part of the spectrum.

I In the visible part the curve will be descending, therefore, theshortest visible wavelength will be the strongest. The star will,therefore, appear blue.


I Planck’s formula

I =2πhe2λ−5

ehc

λkT − 1I Planck’s assumption to justify the formula was that

I E = nhf , where n is an integer.

I n is called the quantum number.

I Planck suggested that energy of molecular vibration is only amultiple of a base energy value.

I Called Planck’s quantum hypothesis.


I Example: Photon energy. Calculate the energy of a photon ofblue light λ = 450 nm (in air).

I Approach

E = hf c = f λ ∴ E =hc

λI Answer

6.63x10−34 × 3.0x108

4.5× 10−7= 4.4× 10−19J

or4.4× 10−19

1.6× 10−19= 2.8eV


I Example: Photons from a lightbulb.Esitmate how many visible light photons a 100W light-bulbemits per second. (Assume energy efficiency is 3% i.e 97%energy is lost as heat).

I Approachλ = 500nm

E =hc

λI Answer

3W = 3Js−1 3 =nhc

λ

∴ n =3

hf=

3× 500× 10−9

6.63× 10−34 × 3.0× 108= 8× 1018


I Exercise: Compare a light source of 1000nm (IR) with a lightsource of 100nm (UV).

E = hf =hc

λ=

hc

1000× 10−9

For λ = 1000nm n =6.63× 10−34 × 3.0× 108

1000× 1.6× 10−38

=12.43

1000× 10−12= 12.4× 109

λ = 100nm n = 12.4× 1010 which 10 times more electrons.


Photoeclectric effect

I Video: A simple gold leaf electroscope.

I Video: Lonnie’s lab. UC Berkeley


What is the kinetic energy and the speed of an electron ejectedfrom a sodium surface who’s work function is W0 = 2.28eV when

(a) λ = 410nm

(b) λ = 550nm

I Approach: Find the energy of the photons. E = hf =hc

λIf E >W0 we get electrons. The maximum kinetic energy isE −W0


Answer

I m = 9.11× 10−31kg. E =1

2mv2

E =hc

λ= 3.03 eV

E −W0 = 3.03− 2.28 ∴ v =

√2E

m

=

√2× 0.75× 1.6× 10−19

9.11× 10−31= 0.513× 106 = 5.1× 105ms−1

I (b) No electrons released.Note: If v is close to 0.1c we have to look at the relativisticequation.


Example 2What is the lowest frequency and the longest wavelength neededto emit electrons from sodium?An exercise left for the student.


Energy, mass and momentum

I We know E = hf

I Photons travel at the speed of light.

I Need a relativistic formulae. (p 6= mv for a photon)

p =mv√1− v2

c2

I Since v = c for a photon, p becomes infinite which isobviously wrong.

I Relativistic equation gives E 2 = p2c2 + m2c4

I Setting m = 0 we get E 2 = p2c2 ∴ p =E

c

I Since E = hf for a photon, p =hf

c=

h

λ


I Example. The 1019 photons emitted per sec from the 100Wlightbulb are focused on a piece of black paper and absorbed.

(a) Calculate the momentum of one photon.(b) Estimate the force of these photons on the paper.

I Approach:

p =h

λ. The momentum changes from p to zero. Use

Newton’s second law, F =∆p

∆t.

I Answer

p =h

λ= 6.63.

If N = 1019 photons, F = 10−8 N.


I Sun

I If the number of photons is large, n × 10−8N can be aconsiderable force!

I Solar sail is an example.


I Photosynthesis

I Chlorophyll in plants captures the energy of sunlight tochange CO2 to useful carbohydrate. 9 photons are needed totransform one molecule of CO2 to carbohydrate and O2.Assuming λ = 670 nm (chlorophyll absorbs most strongly inthe range 650nm - 700nm), how efficient is the photosyntheticprocess? The reverse chemical reaction releases an energy of4.9eV/molecule of CO2.

I ApproachEfficiency is the minimum energy required (4.9eV) divided bythe actual energy absorbed ( 9× the energy of one photon).

Answer: hf =hc

λ.

9× hcλ = 2.7× 10−18 J. = 17eV. ∴4.9

17= 29% efficient.


I Compton effect. AH Compton (1892 - 1962)

I λ′ = λ+h

mec(1− cosφ)

I λ′ − λ is the Compton shift.

I Used to measure bone density for osteoperosis.


Photon interactions - Summary We get:

1. Photoelectric effect. Knocks an electron out of an atom.Photon disappears.

2. Excited state of atom (see later). Photon disappears.

3. Scattered electron and photon. Compton effect.

4. Pair production. Creates matter (e.g. e+ and e−). Photondisappears.


Wave nature of electron.

I Louis de Broglie (1892-1987)

I He proposed that the wavelength of a material particle wouldbe related to its momentum (in same way as for a photon).

p =h

λwhich implies λ =

h

pI Note that this is valid classically, p = mv for (v << c) and

relativistically p =mv√

1− v2

c2

.

I λ =h

pis called the de Broglie wavelength of a particle.


Example. Wavelength of a ball.Calculate the de Broglie wavelength of a 0.20 kg ball moving withspeed of 15ms−1.

I Approach

Use λ =h

p.

I Answerλ = 2.2× 10−34m


ExampleWhat is the wavelength of an electron that has been acceleratedthrough a potential of 100V.

I Approach

Kinetic energy =1

2mv2 = eV , λ =

h

mvI Answer λ = 1.2× 10−10 m = 0.12nm.


Electron diffraction

I Assume electrons strike perpendicular to the surface of a solidand their energy is low. K = 100eV.If the smallest angle at which diffraction maximum occurs is24◦, what is the separation between the atoms on the surface.

Quantum theory and models of the atomElectron diffraction

I ApproachTreat electrons as waves. Constructive interference occurswhen the difference in path = λ, that is, a sin θ = λ

K =p2

2me=

h2

2meλ2∴ λ =

√h

2meK= 0.123nm.

∴ a =λ

sin θ=

0.123

sin 24◦= 0.30 nm.


No-one has actually seen an electron but . . .

they are good for looking at small things. The ebola virus attacksa cell. Transmission electron microscope (x 50,000). Scanningelectron microscope (x35,000)


Transmission electron microscope (x 50,000).Scanning electron microscope (x35,000)


Spectra

I Excited gasses in a discharge tube produce discrete spectra.


I The emission spectra is a characteristic of the material andcan serve as a finger print.

I Also if a continuous spectrum passes through a rarefied gas,dark lines appear where the emission lines occurred. This iscalled an absorption spectrum.


Hydrogen is the simplest atom. One electron. Simplest spectrum.

I J.J. Balmer (1825 - 1898) showed that the 4 lines in the

visible spectrum have wavelengths that fit1

λ= R(

1

22− 1

n2)

for n = 3, 4, . . .

I R is the Rydberg constant. R = 1.0974× 107m−1

I Balmer lines extend into the UV region ending at λ = 365nm.

I Later experiments found that if1

22was replaced with

1

12,

1

32,

1

42.


In general

I1

λ= R(

1

n′2− 1

n2)

I n′ = 1 for Lyman series.

I n′ = 2 for Balmer series.

I n′ = 3 for Paschen series.


The Bohr model

I Electrons in orbits.

I Angular momentum = L = mvrn =nh

2πfor n = 1, 2, 3, . . .

I F =1

4πε0

(Ze)e

r2Coulomb’s law.

I Newton’s second law gives F = ma.

I The acceleration of the particle is a =v2

rn

I ∴1

4πε0

Ze2

rn2=

mv2

rn

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Quantum theory and models of the atomeggnchips.bucket.s3-website-eu-west-1.amazonaws.com/.../Photoelectric2.pdf · Quantum theory and models of the atom I Radiation intensity is proportional