Quantum Physics and Nuclear Physics. 13.1 Quantum Physics. Quantum Physics Revision When an electron falls to a lower energy level , the change in energy ( Δ E) is emitted as a photon of em radiation. . Δ E = E 2 – E 1
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13.1 Quantum Physics
E = hf
Thus the emission spectrum for Hydrogen (this shows only the visible section)...
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Electrons will only be emitted from zinc by photoelectric emission if the electromagnetic radiation incident upon its surface has a frequency of 1 x 1015 Hz or above. This is called the threshold frequency of zinc.
Limitation of Wave Theory of Light
Wave theory would suggest that once enough visible light energy had been absorbed by the zinc, the electron would be able to escape. This is not the case. No matter how intense the incident radiation, if its frequency is below the threshold frequency for a particular material, no photoelectric emission will occur.
E = hf
When uv light is incident upon a zinc surface, each photon gives its energy to a single electron on the zinc surface. uv photons have a high frequency. As a result they give enough energy to the electron to escape from the surface.
The minimum energy needed to just remove an electron from a metal surface is called the work function, .
If the incident light has a lower frequency, each photon has less energy and so no electrons are emitted, irrespective of the intensity.
If the photon energy (E=hf) is greater than the work function (), any remaining energy becomes kinetic energy of the electron (= ½mv2). Thus Einstein stated...
This is one version of Einstein's photoelectric equation.
hf = + ½ mv2
m = mass of an electron
v = speed of fastest electrons (ms1)
Thus, if the p.d. applied was known, the KE ‘removed’ from the fastest electron can be found:
We know... V = W / q W = eV
So...
Link  PhET simulation
V = stopping voltage
e = charge on an electron
= 1.6 x 1019C
= Work function (Joules)
hf = + eV
V = h f 
e e
Q. Explain how you would determine a value for the work function of the metal used to produce the graph above.
Q.
If the metal surface and frequency of incident radiation are both kept constant, a graph can be plotted showing how the photoelectric current (photocurrent) in a photocell varies with applied p.d. (voltage).
Consider these situations and explain the photocurrent that will flow in each case:
+


+
We have seen that light behaves both like a wave (it diffracts) and like a particle (in explaining photoelectric emission). It has wave particle duality.
Demo 1: Shine a beam of laser light (a wave) through a single diffraction grating (like a very fine gauze) then through many gratings crossed at angles to each other.
Demo 2: Fire a beam of electrons (particles of matter) at a thin piece of graphite using a cathode ray tube.
Wave  particle duality is the ability of something to exhibit both wave and particle behaviour.
Considering a photon of light, in 1924 Prince Louis Victor de Broglie equated Einstein’s massenergy relation and Planck’s equation:
E = mc2 and E = hf
thus... mc2 = hf
so... mc2 = h c
λ
so... or
But c = f λ
λ = h
mc
λ = h
p
( Where h = Planck’s constant = 6.6 x 1034Js )
By analogy, this equation can be applied to any other particle of matter. Thus de Broglie’s Hypothesis:
Q1 A year 13 student runs with joy to his physics lesson. If he runs at 5 ms1 and has mass 60kg, determine the de Broglie wavelength of his motion. Comment upon your answer.
All matter can behave like a wave, with the wavelength given by Plank’s constant divided by the matter’s momentum.
Atoms of a gas emit em radiation if they become excited. This means the electrons jump to a higher energy level and then fall back, losing potential energy and emitting it as a photon of em radiation. This will have a frequency according to ΔE = hf
Conclusion
Explain why...
only certain colours are seen for any particular lamp.
coloured fringes are produced
This is the image seen when light from one lamp is diffracted through a grating:
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If light with a continuous spectrum of frequencies (a filament light bulb approximately emits this) is shone into a gas its photons will interact with electrons in the gas atoms, boosting them to a higher energy level. Thus the gas will absorb only certain frequencies of the light. These frequencies will be missing from the light that passes through.
Q. Why des the intensity not fall to zero for the absorption lines?
Energy is reemitted as photons in all directions
This effect is visible in the Fraunhoffer lines seen in spectra of light from the Sun. The Sun emits a virtually continuous spectrum. The absorption lines are due to gases in the Sun’s atmosphere absorbing some frequencies of photon. Hence astronomers can deduce what gases are in a star’s atmosphere as well as their proportions.
We have seen how the Bohr model of an atom suggests that the atom has discreet energy levels. Thus the energy of the electrons and of the atom is not continuous.
Next we will look at how this idea is compatible with the de Broglie hypothesis.
According to de Broglie, the electron must have wave properties.
So how does this wave fit into an atom...?
Clearly, if there are nodes at either end, the electron can only have certain fixed wavelengths and frequencies.
i.e. λ = 2L (fundamental)
λ = L (2nd harmonic)
λ = 2/3 L (3rd harmonic)
So in general λ = 2L
n
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The KE of the electron is given by KE = ½ mv2
Substituting the equations for v and λ gives…
The energy of the electron ‘wave’ is clearly quantised (it has discreet values depending upon n), thus fitting scientific observations e.g. work function in the photoelectric effect.
λ = h = h v = h
p mv mλ
n = integer
h, m and L are all constants.
KE = n2 h2
8mL2
The Bohr model of the atom worked well with basic ideas of quantum theory (as we have seen). However it has limitations.
E.g. It only works well with the Hydrogen atom
It does not predict intensities of different spectral lines.
This illustrates an important feature of quantum mechanics. Its outcomes are probabilities and not certainties; it is a probabilistic model.
Classical mechanics (e.g. Newtons laws) are deterministic: the future is determined by the past and is therefore predictable with some certainty!
Consider one electron amongst a beam of electrons moving with identical momentum towards a narrow slit:
If the slit is wide, the electron will pass straight through without diffracting. Hence we can know its direction and thus its momentum beyond the slit. However we cannot be sure where exactly it passed through the slit so the uncertainty in position is large.
So we are not sure about its direction and thus momentum although we can be quite sure of its position.
This is an example of the Heisenberg uncertainty principle which states that...
Note: This is a fundamental property of the universe and is nothing to do with our ability to measure accurately.
It is not possible to measure the exact position and the exact momentum of a particle at the same time.
This can also be written in terms of energy and time:
Clearly the value of h is so small that the effects of this uncertainty are not seen in everyday events.
h = Plank’s constant
Δp Δx ≥ h
4π
ΔE Δt ≥ h
4π
Consider a football moving through the air. If the uncertainty in its momentum is 2.25 x 102 kgms1, Determine the uncertainty in its position (Δx).
Δx ≥ +/ 2.33x1033m
Δx Δp ≥ h
4π
so… Δx ≥ h
4πΔp
Δx ≥ 6.6 x 1034
4π x (2.25 x 102)
A proton in the LHC takes 1.0 x 1010s to collide. Assuming a 1% uncertainty in measurement of time, determine the uncertainly in measurements of proton energy (ΔE).
1% uncertainty in time = 1/100 x (1.0 x 1010)
= +/ 1.0 x 1012 s
ΔE Δt ≥ h
4π
so… ΔE ≥ h
4πΔt
ΔE ≥ 6.6 x 1034
4π x (1.0 x 1012)
ΔE ≥ +/ 5.3 x 1023 J
Assuming 1% uncertainty in velocity, determine the uncertainty in the position of a proton moving at
2.5 x 107ms1
Close to the speed of light (3.0 x 108 ms1)
( mp = 1.673 x 1027 kg )
Outline by reference to position and momentum how the Schrödinger model of the hydrogen atom is consistent with the Heisenberg uncertainty principle.
 the Schrödinger model assigns a wave function to the electron that is a measure of the probability of finding it somewhere; therefore the position of the electron is uncertain; resulting in an uncertainty in its momentum;
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From the de Broglie hypothesis we can see that our picture of the electron as a small ‘ball’ is too simplistic. It
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