24.1: Quantum Mechanics
Quantum Mechanics, discovered early in the
Electromagnetic radiation is carried by particles, called photons, which interact with electrons. Depending on the experiment, photons can behave as particles or waves. The reverse is also true; electrons can also behave as particles or waves.
Because the electron has a wavelength, its position and momentum can never be precisely established. This is called the uncertainty principle. (What has been said above about the electron is true for protons or any other particle, but, experimentally, the effects become undetectable with increasing mass.)
The Key Concepts
 The energy of a photon is the product of its frequency and Planck’s Constant. This is the exact amount of energy an electron will have if it absorbs a photon.
 A photon, which has neither mass nor volume, carries energy and momentum; the quantity of either energy or momentum in a photon depends on its frequency. The photon travels at the speed of light.
 The five conservation laws hold true at the quantum level. Energy, momentum, angular momentum, charge and CPT are all conserved from the particle level to the astrophysics level.
 If an electron loses energy the photon emitted will have its frequency (and wavelength) determined by the difference in the electron’s energy. This obeys the conservation of energy, one of the five conservation laws.
 An electron, which has mass (but probably no volume) has energy and momentum determined by its speed, which is always less than that of light. The electron has a wavelength determined by its momentum.
 If a photon strikes some photoelectric material its energy must first go into releasing the electron from the material (This is called the work function of the material.) The remaining energy, if any, goes into kinetic energy of the electron and the voltage of an electric circuit can be calculated from this. The current comes from the number of electrons/second and that corresponds exactly to the number of photons/second.
 Increasing the number of photons will not change the amount of energy an electron will have, but will increase the number of electrons emitted.
 The momentum of photons is equal to Planck’s constant divided by the wavelength.
 The wavelength of electrons is equal to Planck’s constant divided by the electron’s momentum. If an electron is traveling at about
.1c this wavelength is then not much smaller than the size of an atom.  The size of the electron’s wavelength determines the possible energy levels in an atom. These are negative energies since the electron is said to have zero potential energy when it is ionized. The lowest energy level (ground state) for hydrogen is
−13.6eV . The second level is−3.4eV . Atoms with multiple electrons have multiple sets of energy levels. (And energy levels are different for partially ionized atoms.)  When an electron absorbs a photon it moves to higher energy level, depending on the energy of the photon. If a
13.6eV photon hits a hydrogen atom it ionizes that atom. If a10.2eV photon strikes hydrogen the electron is moved to the next level.  Atomic spectra are unique to each element. They are seen when electrons drop from a higher energy level to a lower one. For example when an electron drops from
−3.4eV to−13.6eV in the Hydrogen atom a10.2eV photon is emitted. The spectra can be in infrared, visible light, ultraviolet and evenX− rays. (The10.2eV photon is ultraviolet.)  The wave nature of electrons makes it impossible to determine exactly both its momentum and position. The product of the two uncertainties is on the order of Plank’s Constant. (Uncertainty in the electron’s energy and time are likewise related.)
The Key Equations
Relates energy of a photon to its frequency.
Relates the momentum of a photon to its wavelength.
The Debroglie wavelength of an electron.
This is the Heisenberg Uncertainty Principle, (HUP) which relates the uncertainty in the momentum and position of a particle.
Relates the uncertainty in measuring the energy of a particle and the time it takes to do the measurement.
Planck’s constant.
The most convenient unit of energy at the atomic scale is the electron volt, defined as the potential energy of the charge of an electron across a potential difference of
A photon of energy of
Problems Set: Quantum Mechanics
 Calculate the energy and momentum of photons with the following frequency:
 From an
FM station at101.9MHz  Infrared radiation at
0.90×1014Hz  From an
AM station at740kHz
 From an
 Find the energy and momentum of photons with a wavelength:
 red light at
640nm  ultraviolet light at
98.0nm  gamma rays at
.248pm
 red light at
 Given the energy of the following particles find the wavelength of:
 Xray photons at
15.0keV  Gamma ray photons from sodium
24 at2.70MeV  A
1.70eV electron
 Xray photons at
 The momentum of an electron is measured to an accuracy of
5.10×10−15kg−m/s . What is the corresponding uncertainty in the position of the electron?  The four lowest energy levels in electronvolts in a hypothetical atom are respectively:
−34eV, −17eV, −3.5eV, −.27eV . Find the wavelength of the photon that can ionize this atom.
 Is this visible light? Why?
 If an electron is excited to the fourth level what are the wavelengths of all possible transitions? Which are visible?
 Light with a wavelength of
620nm strikes a photoelectric surface with a work function of1.20eV . What is the stopping potential for the electron?  For the same surface in the previous problem but different frequency light, a stopping potential of
1.40V is observed. What is the wavelength of the light?  An electron is accelerated through
5000V . It collides with a positron of the same energy. All energy goes to produce a gamma ray. What is the wavelength of the gamma ray ignoring the rest mass of the electron and positron?
 Now calculate the contribution to the wavelength of the gamma ray of the masses of the particles? Recalculate the wavelength.
 Was it safe to ignore their masses? Why or why not?
 An photon of
42.0eV strikes an electron. What is the increase in speed of the electron assuming all the photon’s momentum goes to the electron?  A
22.0keV X− ray in thex direction strikes an electron initially at rest. This time a0.1nm X− ray is observed moving in thex− direction after collision. What is the magnitude and direction of the velocity of the electron after collision?  The highly radioactive isotope Polonium \begin{align*}214\end{align*}
214 has a halflife of \begin{align*}163.7 \mu s\end{align*}163.7μs and emits a \begin{align*}799 \;\mathrm{keV}\end{align*}799keV gamma ray upon decay. The isotopic mass is \begin{align*}213.99 \;\mathrm{amu}\end{align*}213.99amu . How much time would it take for \begin{align*}7/8\end{align*}
7/8 of this substance to decay?  Suppose you had \begin{align*}1.00 \;\mathrm{g}\end{align*}
1.00g of \begin{align*}Po^{214}\end{align*}Po214 how much energy would the emitted gamma rays give off while \begin{align*}7/8\end{align*}7/8 decayed?  What is the power generated in kilowatts?
 What is the wavelength of the gamma ray?
 How much time would it take for \begin{align*}7/8\end{align*}
 Ultraviolet light of \begin{align*}110 \;\mathrm{nm}\end{align*}
110nm strikes a photoelectric surface and requires a stopping potential of \begin{align*}8.00\end{align*}8.00 volts. What is the work function of the surface?  Students doing an experiment to determine the value of Planck’s constant shined light from a variety of lasers on a photoelectric surface with an unknown work function and measured the stopping voltage. Their data is summarized below:
 Construct a graph of energy vs. frequency of emitted electrons.
 Use the graph to determine the experimental value of Planck’s constant
 Use the graph to determine the work function of the surface
 Use the graph to determine what wavelength of light would require a \begin{align*}6.0 \;\mathrm{V}\end{align*}
6.0V stopping potential.  Use the graph to determine the stopping potential required if \begin{align*}550 \;\mathrm{nm}\end{align*}
550nm light were shined on the surface.
Laser 
Wavelength \begin{align*}(nm)\end{align*} 
Voltage \begin{align*}(V)\end{align*} 

HeliumNeon 
\begin{align*}632.5\end{align*} 
\begin{align*}.50\end{align*} 
KryptonFlouride 
\begin{align*}248\end{align*} 
\begin{align*}3.5\end{align*} 
Argon 
\begin{align*}488\end{align*} 
\begin{align*}1.1\end{align*} 
Europium 
\begin{align*}612\end{align*} 
\begin{align*}.60\end{align*} 
Gallium arsenide 
\begin{align*}820\end{align*} 
\begin{align*}.05\end{align*} 
 An element has the following six lowest energy (in \begin{align*}eV\end{align*}
eV ) levels for its outermost electron: \begin{align*} 24 \;\mathrm{eV},\end{align*}−24eV, \begin{align*}7.5 \;\mathrm{eV},\end{align*}−7.5eV, \begin{align*}2.1 \;\mathrm{eV},\end{align*}−2.1eV, \begin{align*}1.5 \;\mathrm{eV},\end{align*}−1.5eV, \begin{align*}.92\;\mathrm{eV},\end{align*}−.92eV, \begin{align*}.69 \;\mathrm{eV}.\end{align*}−.69eV.  Construct a diagram showing the energy levels for this situation.
 Show all possible transitions; how many are there?
 Calculate the wavelengths for transitions to the \begin{align*}7.5 \;\mathrm{eV}\end{align*}
−7.5eV level  Arrange these to predict which would be seen by infrared, visible and ultraviolet spectroscopes
 A different element has black absorption lines at \begin{align*}128 \ \mathrm{nm}, 325 \ \mathrm{nm}, 541 \ \mathrm{nm} \ \mathrm{and} \ 677 \ \mathrm{nm}\end{align*}
128 nm,325 nm,541 nm and 677 nm when white light is shined upon it. Use this information to construct an energy level diagram.  An electron is accelerated through \begin{align*}7500 \;\mathrm{V}\end{align*}
7500V and is beamed through a diffraction grating, which has \begin{align*}2.00 \times 10^7\end{align*}2.00×107 lines per \begin{align*}cm\end{align*}cm . Calculate the speed of the electron
 Calculate the wavelength of the electron
 Calculate the angle in which the first order maximum makes with the diffraction grating
 If the screen is \begin{align*}2.00 \;\mathrm{m}\end{align*}
2.00m away from the diffraction grating what is the separation distance of the central maximum to the first order?
 A light source of \begin{align*}429 \;\mathrm{nm}\end{align*}
429nm is used to power a photovoltaic cell with a work function of \begin{align*}0.900 \;\mathrm{eV}\end{align*}0.900eV . The cell is struck by \begin{align*}1.00 \times 10^{19}\end{align*}1.00×1019 photons per second. What voltage is produced by the cell?
 What current is produced by the cell?
 What is the cell’s internal resistance?
 A \begin{align*}.150 \;\mathrm{nm}\end{align*}
.150nm \begin{align*}X\end{align*}X− ray moving in the positive \begin{align*}x\end{align*}x− direction strikes an electron, which is at rest. After the collision an \begin{align*}X\end{align*}X− ray of \begin{align*}0 .400 \;\mathrm{nm}\end{align*}0.400nm is observed to move \begin{align*}45\end{align*}45 degrees from the positive \begin{align*}x\end{align*}x− axis. What is the initial momentum of the incident \begin{align*}X\end{align*}
X− ray?  What are the \begin{align*}x\end{align*}
x and \begin{align*}y\end{align*}y components of the secondary \begin{align*}X\end{align*}X− ray?  What must be the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} components of the electron after collision?
 Give the magnitude and direction of the electrons’ final velocity.
 What is the initial momentum of the incident \begin{align*}X\end{align*}
 Curium \begin{align*}242\end{align*} has an isotopic mass of \begin{align*}242.058831 \;\mathrm{amu}\end{align*} and decays by alpha emission; the alpha particle has a mass of \begin{align*}4.002602 \;\mathrm{amu}\end{align*} and has a kinetic energy of \begin{align*}6.1127 \;\mathrm{Mev}\end{align*}.
 What is the momentum of the alpha particle?
 What is its wavelength?
 Write a balanced nuclear equation for the reaction.
 Calculate the isotopic mass of the product.
 If the alpha particle is placed in a magnetic field of \begin{align*}.002 \;\mathrm{T}\end{align*} what is the radius of curvature? (The alpha particle has a double positive charge.)
 If the alpha particle is moving in the \begin{align*}x\end{align*}direction and the field is in the \begin{align*}z\end{align*}direction find the direction of the magnetic force.
 Calculate the magnitude and direction of the electric field necessary to make the alpha particle move in a straight line.
 A student lab group has a laser of unknown wavelength, a laser of known wavelength, a photoelectric cell of unknown work function, a voltmeter and test leads, and access to a supply of resistors.
 Design an experiment to measure the work function of the cell, and the wavelength of the unknown laser. Give a complete procedure and draw an appropriate circuit diagram. Give sample equations and graphs if necessary.
 Under what circumstances would it be impossible to measure the wavelength of the unknown laser?
 How could one using this apparatus also measure the intensity of the laser (number of photons emitted/second)?
 The momentum of an electron is measured to an accuracy of \begin{align*}\pm 5.1 \times 10^{24} \;\mathrm{kg} \cdot \;\mathrm{m/s}\end{align*}. What is the corresponding uncertainty in the position of the same electron at the same moment? Express your answer in Angstroms (\begin{align*}1 \AA = 10^{10} \;\mathrm{m}\end{align*}, about the size of a typical atom).
 Thor, a baseball player, passes on a pitch clocked at a speed of \begin{align*}45 \pm 2 \;\mathrm{m/s}\end{align*}. The umpire calls a strike, but Thor claims that the uncertainty in the position of the baseball was so high that Heisenberg’s uncertainty principle dictates the ball could have been out of the strike zone. What is the uncertainty in position for this baseball? A typical baseball has a mass of \begin{align*}0.15 \;\mathrm{kg}\end{align*}. Should the umpire rethink his decision?
 Consider a box of empty space (vacuum) that contains nothing, and has total energy \begin{align*}\;\mathrm{E} = 0\end{align*}. Suddenly, in seeming violation of the law of conservation of energy, an electron and a positron (the antiparticle of the electron) burst into existence. Both the electron and positron have the same mass, \begin{align*}9.11 \times 10^{31} \;\mathrm{kg}\end{align*}.
 Use Einstein’s formula \begin{align*}(\;\mathrm{E = mc}^2)\end{align*} to determine how much energy must be used to create these two particles out of nothing.
 You don’t get to violate the law of conservation of energy forever – you can only do so as long as the violation is “hidden” within the HUP. Use the HUP to calculate how long (in seconds) the two particles can exist before they wink out of existence.
 Now let’s assume they are both traveling at a speed of \begin{align*}0.1 \;\mathrm{c}\end{align*}. (Do a nonrelativistic calculation.) How far can they travel in that time? How does this distance compare to the size of an atom?
 What if, instead of an electron and a positron pair, you got a proton/antiproton pair? The mass of a proton is about \begin{align*}2000 \times\end{align*} higher than the mass of an electron. Will your proton/antiproton pair last a longer or shorter amount of time than the electron/positron pair? Why?
Answers to Selected Problems

 \begin{align*}6.752 \times 10^{26} J, 2.253 \times 10^{34} \;\mathrm{kgm/s}\end{align*}
 \begin{align*}5.96 \times 10^{20} J, 1.99 \times 10^{28} \;\mathrm{kgm/s}\end{align*}
 \begin{align*}4.90 \times 10^{28} J, 1.63 \times 10^{36} \;\mathrm{kgm/s}\end{align*}
 \begin{align*}1.94 \;\mathrm{eV}, 1.04 \times 10^{27} \;\mathrm{kgm/s}\end{align*}
 \begin{align*}12.7 \;\mathrm{eV}, 6.76 \times 10^{27} \;\mathrm{kgm/s}\end{align*}
 \begin{align*}5.00 \;\mathrm{eV}, 2.67 \times 10^{21} \;\mathrm{kgm/s}\end{align*}
 \begin{align*}.0827 \;\mathrm{nm}\end{align*}
 \begin{align*}4.59 \times 10^{4}\;\mathrm{nm}\end{align*}
 \begin{align*}.942 \;\mathrm{nm}\end{align*}

\begin{align*}1.03 \times 10^{20}\;\mathrm{m}\end{align*}
 \begin{align*}36 \;\mathrm{nm}\end{align*}
 no
 \begin{align*}380 \;\mathrm{nm}, 73 \;\mathrm{nm}, 36 \;\mathrm{nm}, 92 \;\mathrm{nm}, 39 \;\mathrm{nm}\end{align*}
 \begin{align*}.80 \;\mathrm{V}\end{align*}

\begin{align*}480 \;\mathrm{nm}\end{align*}
 \begin{align*}.124 \;\mathrm{nm}\end{align*}
 \begin{align*}.00120 \;\mathrm{nm}\end{align*}
 \begin{align*}24,600 \;\mathrm{m/s}\end{align*}

\begin{align*}1.84 \times 10^8\;\mathrm{m/s}\end{align*}
 \begin{align*} .491 \;\mathrm{m/s}\end{align*}
 \begin{align*}3.14 10^7\;\mathrm{J}\end{align*}
 \begin{align*}64 \;\mathrm{Mw}\end{align*}
 \begin{align*}1.55 \;\mathrm{pm}\end{align*}
 \begin{align*}3.27 \;\mathrm{eV}\end{align*}
 .
 b. \begin{align*}15\end{align*} c. \begin{align*}182 \;\mathrm{nm}, 188 \;\mathrm{nm}, 206 \;\mathrm{nm}, 230 \;\mathrm{nm}\end{align*}

\begin{align*}10.3 \;\mathrm{eV}, 3.82 \;\mathrm{eV}, 2.29 \;\mathrm{eV}, 1.83 \;\mathrm{eV}\end{align*}
 \begin{align*}4.19 \times 10^7\;\mathrm{m/s}\end{align*}
 \begin{align*}1.70 \times 10^{11}\;\mathrm{m}\end{align*}
 \begin{align*}1.95^\circ\end{align*}
 \begin{align*}.068 \;\mathrm{m}\end{align*}
 \begin{align*}1.89 \;\mathrm{V}\end{align*}
 \begin{align*}1.60 \;\mathrm{A}\end{align*}
 \begin{align*}1.25 \ \Omega\end{align*}
 \begin{align*}4.40 \times 10^{24}\;\mathrm{kgm/s}\end{align*}
 \begin{align*}1.17 \times 10^{24}\;\mathrm{kgm/s}\end{align*}
 \begin{align*}3.23 \times 10^{24}\;\mathrm{kgm/s}\end{align*}
 \begin{align*}3.76 \times 10^7\;\mathrm{m/s}\end{align*}
 \begin{align*}1.1365 \times 10^{22}\;\mathrm{kgm/s}\end{align*}
 \begin{align*}5.860 \;\mathrm{pm}\end{align*}
 \begin{align*}^242\;\mathrm{Cu} \rightarrow ^4\mathrm{He} + ^238\mathrm{Pu}\end{align*}
 \begin{align*}238.0497 \;\mathrm{amu}\end{align*}
 \begin{align*}17.7 \;\mathrm{cm}\end{align*}
 \begin{align*}y\end{align*}
 \begin{align*}+y, 34.2 \;\mathrm{N/C}\end{align*}
 .
 0.10 Angstrom
 \begin{align*}1.76 \times 10^{34} \;\mathrm{eV}\end{align*}