Chapter 27 Early Quantum Theory and Models of the Atom 27-1 Discovery and Properties of the electron Measure charge to mass ratio e/m (J. J. Thomson, 1897)

• When apply magnetic field only, the rays are deflected downward and bent into a curved path (path c)

FB = evB F = ma → em=vBr

The magnetic field B and the radius of curvature r can be measured.

• When electric field and magnetic field are applied together, and E is adjusted so that the cathode rays are undeflected (path b):

FB = FE, evB = eE → v= E/B The charge to mass ration

em=

EB2r

=1.76 × 1011 C/kg

Measure e: Oil-drop experiment (R. A. Millikan, 1897) • Tiny oil droplets carrying an

electric charge fall under gravity.

• Electric field E between the

plates was adjusted until the oil drop was suspended in midair. q=mg/E

• The mass of the droplet was determined by measuring its terminal velocity without the electric field

• After hundreds painstaking observations and analysis, Millikan

discovered that any electric charge was an integral multiple of a smallest charge, e = 1.6 × 10-19 C. Electric charge is quantized.

• With measurement of both e and e/m, the mass of the electron

= 9.1 × 10-31 kg. This mass is less than 1/1000 the mass of the smallest atom, thus confirmed that the electron is only a part of an atom.

27-2 Planck's Quantum Hypothesis • Blackbody is a body that would absorb all the radiation falling

on it. It would appear black. • Blackbody radiation: The EM radiation a

blackbody would emit when hot and luminous. • As the temperature increases, the EM radiation

is the strongest at higher frequencies. Wien's law: λpT = 2.90 × 10-3 m•K. λp is the wavelength at the peak of the spectrum. T is temperature in Kelvin.

Example: Estimate the temperature of the surface of our Sun, given that the Sun emits light whose peak is in the visible spectrum at around 500 nm. Use Wien's law:

T = 2.90 ×10−3m ⋅K

λp=2.90 ×10−3m ⋅K500 ×10−9m

≈ 6000K

Planck's quantum hypothesis:

Max Plank (1900) showed that the blackbody radiation curve can be explained if molecular oscillations are quantized: their energy E can only be integer (n) multiples of hf:

E = nhf (n = 1, 2, 3, …) h = 6.626 × 10-34 J•s, is called Planck's constant. f is the frequency of oscillation.

The smallest amount of energy possible Emin = hf. 27-3 Photon Theory of Light and the Photoelectric Effect Einstein's photon theory (1905): Since the vibrational energy of molecules in a radiating object is quantized with energy E = nhf, when light is emitted by a molecular oscillator, its energy must decrease by an amount fh (or 2hf, …) to another integer × hf. Then to conserve energy, the light ought to be emitted in packet or quanta, each with an energy E = hf. Light is transmitted as tiny particles, or photon.

• Photoelectric effect: when light shines on a metal surface,

electrons are emitted from the surface

• Einstein predicted that in photoelectric effect experiments, an electron is ejected from the metal by a collision with a single photon. All the photon energy is transferred to the electron. A minimum energy (work function) W0 is required to get an electron out through the surface. His theory predicted:

1) An increase in intensity of the light beam means more photon

are incident, so more electrons will be ejected; but since the energy of each photon is not changed, the maximum KE if electron is not changed.

2) If the frequency of the light is increased, the maximum KE of the electrons increases linearly:

KEmax = hf - W0

3) If the frequency f is less than the "cutoff" frequency f0, where hf0 = W0, no electrons will be ejected at all, no matter how great the intensity.

• The experimental results ( R. A.

Millikan, 1914) were fully in agreement with Einstein's photon theory.

Example: Calculate the energy of a photon of blue light,

λ = 450 nm.

E = hf = hcλ=(6.63 ×10−34 J • s)(3.0 ×108m / s)

450 ×10−9m= 4.4 ×10−19 J

= (4.4 ×10−19 J ) /(1.6×10−19 J / eV)= 2.7eV

Example: Estimate how many visible light photon a 100W lightbulb emits per second. Assume the average wavelength = 500 nm.

The energy emitted in 1 second = 100 J. Let n be the number of photon emitted per second, then

Frequency of light f f0

KEmax

n = E

hf=Eλhc

=(100J )(500×10−9m)

(6.63×10−34 J • s)(3.0×108m / s)

= 2.5×1020

Example: What is the maximum kinetic energy of an electron ejected from a sodium surface whose work function is W0 =2.28 eV when illuminated by light of 410 nm.

At λ = 410 nm, E = hf = hcλ=(6.63 ×10−34 J • s)(3.0 ×108m / s)

410 ×10−9m

= 4.85×10−19 J = 3.03eV KEmax = hf - W0 = 3.03eV - 2.28eV = 0.75eV or 1.2× 10-19 J. Stefan-Boltzmann Law:

The total energy radiated per unit surface area of a black body across all wavelengths per unit time, j*, is directly proportional to the fourth power of the black body's thermodynamic temperature T:

j* = σT4.

σ = 2π5k4/(15c2h3) = 5.670373 × 10-8 W m-2 K-4 is the Stefan-Boltzmann constant.

27-4 Compton Effect and Pair Production Compton Effect (A. H. Compton, 1923): In X-ray scattering, the scattered light had a slightly lower frequency (longer wavelength) than did the incident light. • Explanation: As incident

photons colliding with electrons of the material, some energy is transferred to the electrons. The energy of photon (i.e., frequency) is lower after the collision.

• Using Einstein's photon theory and laws of conservation of

energy and momentum, he derived:

λ'= λ + hm0c

(1− cosφ)

m0 is the rest mass of an electron. h/m0c is called Compton wavelength. • The wave theory predicts no change in wavelength of

scattered light. • Compton effect confirmed the photon theory of light.

Pair production:

A photon disappears and an electron-position pair is

created. E = mc2 The reverse process:

If an electron collides with a positron, the two annihilate each other and their energy including their mass converted to the energy of the photon. Q: Can a photon disappear and produce two electrons? Example: (a) What is the minimum energy of a photon that can produce

an electron-positron pair? E = mc2 = 2mec

2 = 2(9.11×10−31kg)(3×108m / s)2

= 1.64 ×10−13 J = 1.02MeV

(b) What is this photon's wavelength?

λ =hcE=(6.6 ×10−34 J • s)(3×108m / s)

(1.64×10−13 J)= 1.2×10−12m

At this wavelength, the EM radiation is called gamma-ray. A photon can interacts with atoms/electrons in following ways: 1. The photon can be scattered off an electron (or a nucleus) and

in the process lose some energy → Compton Effect. 2. A photon may knock an electron out of an atom and in the

process itself disappear → Photoelectric effect. 3. The photon may knock an atomic electron to a higher energy

state in the atom if its energy is not sufficient to knock the electron out altogether. The photon disappears, and all its energy is given to the atom. Such an atom is then in an excited state.

4. A photon can actually create matter, such as the production of

an electron and a position. 27-5 Wave-Particle Duality; the Principle of Complementarity Wave-particle duality: in some experiment, light behaves like a wave, in others it behaves like a stream of particle.

Principle of complementarity: to understand any given experiment, we must use either the wave or the photon theory, but not both. 27-6 Wave Nature of Matter L. de Broglie (1923) proposed that any particle has wave-particle duality. The wavelength of a material particle of mass m traveling with speed v, is given by

λ =hmv

λ is called the de Broglie wavelength of a particle. Example: Calculate the de Broglie wavelength of a 0.20 kg ball

moving with a speed of 15 m/s.

λ =hmv

=(6.6 ×10−34 J .s)(0.20kg)(15m / s)

= 2.2×10−34m

• The wave properties such as interference or diffraction can

only be detected when the size of objects or slits is similar to the wavelength. The size of the smallest atom is about 10-10 m. The wavelength of any ordinary object is much too small to detect.

Example: Calculate the wavelength of an electron that has been

accelerated through a potential difference of 100V.

12mv

2 = eV

v = 2eV /m = 2(1.6×10−19C)(100V) /(9.1×10−31kg)

= 5.9 ×106m / s

λ =hmv

=(6.6 ×10−34 J.s)

(9.1×10−31kg)(5.9 ×106m / s)= 1.2×10−10m

• Since the spacing of atoms

in a crystal is on the order of 10-10 m, they can be used as a diffraction grating for electron beam. The diffraction patterns of electron is observed.

• This wave property of

electron let to development of electron microscope, which can produce images of much greater

magnification than a light microscope. • Protons, neutrons, and other particles also can be used to

generate diffraction pattern.

27-9 Atomic Spectrum: Key to the Structure of the atom Emission Spectrum: • Emission spectrums of low-

density atomic gases were observed in a “discharge tube”.

• The spectrums were discrete, not continuous.

• An emission spectrum is unique for a particular gas, it can serves as a “fingerprint” to identify the gas.

Absorption Spectrum: • When a continuous spectrum passes through a gas, dark lines are observed in the emerging

spectrum, at wavelength corresponding to lines normally emitted by the gas.

• Gases can absorb light at the same frequency at which they emit

Spectrums of Hydrogen Atom • Hydrogen is the simplest atom. It has only one electron

orbiting it nucleus. • J. J. Balmer (1885) showed that the four visible lines in the

hydrogen spectrum (656 nm, 486nm, 434nm, and 410nm) fit the following formula (Balmer series)

1λ

= R 122

−1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ , n = 3, 4, 5, 6, …

R = 1.097 × 107 m–1 is the Rydberg constant. • Later, two other series of lines were discovered in UV and IR:

Lyman series (UV): 1λ

= R 112−1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ , n = 2, 3, 4, 5, …

Paschen series (IR): 1λ

= R 132−1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ , n = 4, 5, 6, 7, …

Q: Is there a relationship between the spectrum and the structure of the atom? Difficulties of the Rutherford’s model of atom: (1) It predicts that the line spectrum

should be continuous. (2) It predicts that atoms are unstable –

electrons should quickly spiral into the nucleus.

electron

nucleus

27-10 The Bohr Model Bohr combine the Rutherford’s model with newly developing quantum theory. Bohr theory postulated: (1) Electrons can only move about the nucleus in certain orbits

(called stationary states). In each orbit, an electron would have a definite value of energy.

(2) An electron in such a stationary state (orbit) emits no

radiation. (3) The angular momentum L of atomic electrons is quantized

by the rule L = nh2π

, n = 1, 2, 3, …

where n is a quantum number. n =1 state is the ground state. n > 1 states are excited states.

(4) The energy of electron in each stationary state is also

quantized, and is given by

En = −2π 2Z 2e 4mke

2

h21n2

, n = 1, 2, 3,

When n → ∞, E → 0. For free electron, E = 0.

(5) If an electron jumps to a lower state, it

emits a photon whose energy equals the difference in energy between the upper and lower states. hf = Eu − El

Atoms are excited to excited states by collision with other atoms or electrons or by absorption of a photon of just the right frequency.

(6) The radius of the circular orbits is given by

r1 =h2

4π 2mkeZe2 , and rn =

n2

Zr1

Z is the number of positive charge of nucleus. Apply Bohr’s model to hydrogen atom

Hydrogen nucleus has one proton, so Z = 1.

(1) Orbits: r1 = 0.529 × 10-10 m, r2 = 4 r1, r3 = 9 r1, r3 = 16 r1, …

(2) Energy:

The ground state energy E1 = -13.6 eV

Generally, En = −13.6eVn2

(3) Emission/absorption spectrum:

When an electron jump from a higher state n to n’, it emits a photon with energy hf = hc

λ= En − En' , thus

1λ

= −2π 2Z 2e4mke

2

h3c1n' 2

−1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= (1.0974×107 / m) 1n' 2

−1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

Example: Calculate the wavelength of light emitted when a hydrogen atom makes a transition from the n = 6 to the n = 2 energy level according to the Bohr model. n' = 2 and n = 6

1λ

= (1.0974×107 /m) 1n'2

−1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= (1.0974×107 /m) 122

−162

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 2.44×106 /m

So λ = 410 nm. This is the fourth line in the Balmer series. Example: Determine the energy difference of an electron in the ground state and the 1st excited state in a hydrogen atom.

En = −13.6eVn2

, so E1 = −13.6eV and E2 = −3.4eV

∆E = E2 - E1 = 10.2eV

27-11 de Broglie's Hypothesis Applied to Atom

Bohr's theory did not explain why the orbits were quantized. de Broglie (1923) proposed that the wave nature of the electron resulted the quantized angular momentum. • A particle of mass m moving with speed v would have a

wavelength of λ = hmv

• Each electron orbit in an atom is actually a circular standing

wave. They are in the resonant mode.

• Waves with other wavelengths interfere with themselves and their amplitudes quickly drop to zero

• Only waves that persist are those for

which the circumference of the circular orbit contains a whole number of wavelengths.

2πrn = nλ, n = 1, 2, 3, … Substitute λ = h/mv, we get

2πrn =nhmv

,

or angular momentum: mvrn = L =nh2π

This is the quantum condition proposed by Bohr. Thus the quantized orbits and energy states in Bohr's model are due to the wave nature of the electron, and that only resonant "standing" waves can persist.