Quantum Mechanics (and a mix of other stuff!)

Equations and constants

lambda* p T = 2.90*10^(-3)mK. ( Wien's law, short lambda )
E = nhf ( Planck's quantum hypothesis )
h = 6.626*10^-34)Js
p = E/c = h/lambda ( momentum of photon )
lambda = lambda + h(1-cosF)/m_0*c ( Scattered photon wavelength )
h/m_0c = Compton wavelength
lambda = h/p ( de Broglie )

dxdp = dEdt = h-bar = h = h/2pi ( uncertainty principle )
L = sqrt(l(l+1))* h ( Angular momentum relation to orbital quantum number )
L_z = m_l h. ( space quantization )
1/lambda = 2pi^2*Z^2*e^4*m*k^2/(h^3*c) * (1/n'^2-1/n^2) (Bohr formula)

Problem 1
What is the velocity of a beam of electrons that goes undeflected when passing through crossed electric and magnetic fields of magnitude 1.38*10^4V/m and 2.90*10^(-3)T? What is the radius of the electron orbit if the electric field is turned off?

In order for the beam of electrons to go undeflected, the force due to electric field and the magnetic field must equal each other.
F = qE = qvB. Therefore v = E/B, v = 4.76*10^6m/s .
The radius of the electron orbit with the electric field turned off can be found by setting the centripetal force equal to the magnetic force.
F = mv^2/r = qvB. Solving for r: r = mv/qB. r = 0.00933m .

Problem 2
An oil drop whose mass is determined to be 2.8*10^(-15)kg is held at rest between two large plates separated by 0.01m when the potential difference between them is 340V. How many excess electrons does this drop have?

In order for the oil drop to be held at rest between the two large plates, the force due to the electric field must be equal to the weight of the oil drop.
F = mg = qE. With E = V/d, mg = qV/d.
Here we can solve for the charge q. q = mgd/V. q = 8.07*10^(-19)C.
The number of excess electrons must be q/e, which is approximately 5 electrons .

Problem 3
A child's swing has a natural frequency of 0.90Hz. a) What is the separation between possible energy values in joules? b) If the swing reaches a vertical height of 0.45m above its lowest point and has a mass of 20kg including the child, what is the value of the quantum number n? c) What is the fractional change in energy between levels whose quantum numbers are n and n + 1? Would quantization be measurable in this case?

a) The separation between energy values in joules can be found by using the equation E = hf, which gives a quantum of energy. E = 5.9634*10^(-34)J .
b) At a vertical height of 0.45m and with a mass of 20kg, the potential energy can be found by PE = mgh => 88.2J. Dividing this by the E found in part a) the quantum number n is then 1.479*10^35 .
c) The fractional change in energy between levels n and n+1 can be found readily. dE = (n+1)hf nhf. dE = hf, in other words the quantum from part a). The fractional change in energy is then found by dividing by the PE, 88.2J, yielding 6.76*10^(-36) . Since this is so small, it would be immeasurable.

Problem 4
What is the longest wavelength of light that will emit electrons from a metal whose work function is 3.10 eV?

The longest wavelength will be given by solving for lambda in E = hc/lambda.
lambda = hc/E. Making the substitutions for the values, and knowing that 1.602*10^(-19)J = 1 eV,
lambda = 4.00*10^(-7)m.

Problem 5
The work functions for sodium, cesium, copper, and iron are 2.3, 2.1, 4.7 and 4.5 eV respectively. Which of these metals will not emit electrons when visible light shines on it?

Visible light is of the range ~400-700nm.
Solving for the energy of such photons by using the equation E = hc/lambda and making the conversion to eV, the respective energies are 3.10eV 1.77eV. Taking the largest energy, 3.10eV, it is apparent by comparing the work function to that, that copper and iron will not emit electrons.

Problem 6
What is the maximum KE of electrons ejected from barium (Work function = 2.48 eV) when illuminated by white light, lambda = 400 to 700nm?

KE of the electrons can be found by the equation KE = hf W, where W is the work function. We notice from the equation E = hc/lambda that the shortest wavelength will result in the largest energy. Therefore we will use 400nm as our wavelength. Solving for energy and converting to eV, we have 3.10eV as the energy of the light. Subtracting the work function, we have 0.62eV as the maximum KE of the electrons.

Problem 7
The threshold wavelength for emission of electrons from a given surface is 320 nm. What will be the maximum kinetic energy of ejected electrons when the wavelength is changed to a) 250nm, b) 350nm?

Since the threshold wavelength for the surface is 320nm, we can solve for the work function of the surface by using E = hc/lambda. 3.878eV.
a) E of 250nm is 4.963eV. Since KE max is E-W, KE max must be 1.058eV.
b) 350nm is greater than the threshold wavelength so no electrons will be ejected.

Problem 8
When 230-nm light falls on a metal, the current through a photoelectric circuit is brought to 0 at a reverse voltage of 1.64V. What is the work function of the metal?

The information tells us the KE of the electrons that are being ejected from the surface. KE = QV, therefore KE = 1.64eV. The energy of the 230nm light is given by E = hc/lambda, 5.395eV. KE = E W, W = 3.755eV.

Problem 9
What is the momentum and the effective mass of a 0.35-nm X-ray photon?

The momentum of a photon is given by p = h/lambda, 1.89*10^(-24)kg*m/s
The effective mass can be found by dividing by the speed of light, 6.31*10^(-33)kg.

Problem 10
What is the longest wavelength photon that could produce a proton-antiproton pair? Each has a mass of 1.67*10^(-27)kg.

We find the total initial energy by using E = mc^2 => 3.006*10^(-10)J. This will be equal to the energy of the photon emitted since the proton and antiproton annihilate each other. E = hc/lambda, lambda = hc/E. lambda = 6.61*10^(-16)m.

Problem 11
What is the minimum photon energy need to produce a + - - pair? The mass of each is 207 times the mass of the electron. What is the wavelength of such a photon?

To find the minimum photon energy needed to produce a + - - pair we consider the total energy of the + - - pair. E = mc^2, m = 207*9.11*10^(-31)kg, E = 3.394*10^(-11)J
The wavelength of the photon with energy E can be found by lambda = hc/E , 5.856*10^(-15)m.

Problem 12
What is the wavelength of an electron of energy a) 10eV, b) 100eV, c) 1.0keV?

We use de Broglie's equation, lambda = h/p. The stated energy can be used to solve for the velocity of the electron. v = sqrt(2KE/m)

a) v = 1.875*10^(6)m/s, lambda = 3.878*10^(-10)m.
b) v = 5.930*10^(6)m/s, lambda = 1.226*10^(-10)m.
c) v = 1.875*10^(7)m/s, lambda = 3.878*10^(-11)m.

Problem 13
What voltage is needed to produce electron wavelengths of 0.10nm? Assume the electrons are nonrelativistic.

We want to relate voltage to energy first of all. KE = QV, V = KE/Q.

Next, we relate energy to velocity KE = 1/2mv^2. V = mv^2/2Q.
Now we relate this to wavelength lambda = h/mv. Solving for v, v = h/m*lambda.
Now we substitute back into our voltage equation. V = m*(h/m*lambda)^2/2Q.
Substituting the knowns in gives a voltage of 150.4153V

Problem 14
What is the potential energy and the kinetic energy of an electron in the ground state of the hydrogen atom?

The potential energy of an electron in the ground state of hydrogen will have an Energy given by eV or kZe^2/r^2. With Z as one, r as 0.529*10^(-10)m the PE is -0.4366*10^(-17)J or -27.2552eV. The KE of the electron can then be found by E 1 = KE PE, where E 1 is -13.6eV. KE = 13.6552eV.

Problem 15
An excited hydrogen atom could, in principle, have a radius of 1.00mm. What would be the value of n for a Bohr orbit of this size? What would its energy be?

With a radius of 10^(-3)m, we can use the relation r 1 *n^2 = r ­n . r 1 is equal to 0.529*10^(-10)m, so n is equal to ~4348. To find the energy, we divided E 1 by n^2, -13.6eV/n^2 => -7.1944*10^(-7)eV.

Problem 16
At low temperatures, nearly all the atoms in hydrogen gas will be in the ground state. What minimum frequency photon is needed if the photoelectric effect is to be observed?

Ground state is n = 1, 13.6eV. The minimum frequency photon must be 13.6eV. So, we use the equation E = hf, and solve for f: 3.288*10^15Hz.

Problem 17
If a 100-W light bulb emits 3.0% of the input energy as visible light (ave lambda = 550nm) uniformly in all directions, estimate how many photons per second of visible light will strike the pupil (4.0mm diameter) of the eye of an observer 1.0km away.

Power = work/time. 100J/s *3/100% => 3J/s. Now, we want to find how many photons are in 3J. E = hf = hc/lambda. E = 3.6142*10^(-19)J/photon. So, we divide and find that there are going to be 8.3*10^(18)photons/s to be emitted.
Now we want to find the fractional part of the surface area that the eye occupies.
pi*(0.002m)^2 = area of eye. 4pi*(1000m)^2 = total area.
(0.002)^2/4(1000)^2 * 8.3*10^(18) => 8.30*10^(7) photons/s.

Problem 18
In certain of Rutherford 's experiments the alpha particles (mass = 6.64*10^(-27)kg) had a KE of 4.8 MeV. How close could they get to a gold nucleus (charge = +79e)? Ignore the recoil motion of the nucleus.

We want to relate energy to distance. Work = QV. V = kQ/r. Work = kQ^2/r. Solving for r gives us 2.373*10^(-14)m. This gives us the distance between the two bodies, overlapping the radius; we want twice that. Multiply by two to get 4.746*10^(-14)m.

Problem 19
For what maximum KE is a collision between an electron and a hydrogen atom in its ground state definitely elastic?

A hydrogen atom in its ground state has a PE of -13.6eV. A collision is definitely elastic if it is from n = 2 to n = 1, so -3.4 (-13.6) => 10.2eV.

Problem 20
Calculate the ratio of the gravitational to electric force for the electron in a hydrogen atom. Can the gravitation force be safely ignored?
F_C = kQq/r^2 and F_G = GMm/r^2. The ratio is then GMm/kQq, substituting the knowns in, we arrive at 4.396*10^(-40), making it ignorable.

Problem 21
An electron in the n = 2 state of hydrogen remains there on the average about 10^(-8)s before jumping to the n = 1 state. a) Estimate the uncertainty in the energy of the n = 2 state. b) What fraction of the transition energy is this? c) What is the wavelength, and width (in nm), of this line in the spectrum of hydrogen?

a) To find uncertainty, we use dEdt = h-bar. Solving for dE, we arrive at 1.055*10^(-26)J.
b) (-3.4) (-13.6) gives the transition energy 10.2eV. The ratio is then 6.4565*10^(-9).
c) The wavelength of this line in the spectrum of hydrogen is given E = hc/lambda. Lambda = 1.2165*10^(-7)m. To find the width of the line, we use the ratio found in part b and the wavelength just determined. The ratio is the fraction of the uncertainty to the transitional energy. Multiplying the ratio to the wavelength gives the fraction of uncertainty in wavelength. 7.85*10^(-7)nm.

Problem 22
A 12-g bullet leaves a rifle at a speed of 180m/s. a) What is the wavelength of this bullet? b) If the position of the bullet is known to an accuracy of 0.60 cm (radius of barrel), what is the minimum uncertainty in its momentum? c) If the accuracy of the bullet were determined only by the uncertainty principle (an unreasonable assumption), by how much might the bullet miss a pinpoint target 200m away?

a) The wavelength is given by de Broglie's equation, lambda = h/p, 3.0676*10^(-34)m.
b) Uncertainty in momentum is given by dp = h-bar/dx, with dx as 0.60/100m:
c) With the uncertainty in momentum from part b), we can find the uncertainty in velocity by dividing by the mass of the bullet: 1.465*10^(-30)m/s. Now, we can use the equation d = vt. To find the time t, we divide the horizontal displacement by the speed: 1.11s. The displacement is then 1.63*10^(-30)m.

Problem 23
Using the Bohr formula for the radius of an electron orbit, estimate the average distance from the nucleus for an electron in the innermost (n = 1) orbit of a lead atom (Z = 82). Approximately how much energy would be required to remove this innermost electron?

The average distance from the nucleus can be found by relating it to hydrogen's innermost radius, 0.529*10^(-10)m. Divide by Z to get the radius for lead, because that is the only thing that has changed: 6.45*10^(-13)m. The energy to remove this innermost electron can be found by multiplying the energy for hydrogen, 13.6eV by Z^2, 91446.4eV.

Problem 24
An X-ray tube operates at 100kV with a current of 25mA and nearly all the electron energy goes into heat. If the specific heat capacity of the 0.085kg plate is 0.11kcal/kgC what will be the temperature rise per minute if no cooling water is used?

First we find the power, P = IV, P = 2500J/s. Now, we convert it to J/min, 1.5*10^5J/min. We set this equal to the Q_plate/t and solve for dT in Q/t = mCdT/t, dt/min = 3832.5C/min. *NOTE: Make sure you notice the units!

Problem 25
In the so-called vector model of the atom, space quantization of angular momentum vector of magnitude L = sqrt(l(l+1)h-bar is thought of as precessing around the z-axis (like a spinning top or gyroscope) in such a way that the z-component of angular momentum L z = m l h-bar, also stays constant. Calculate the possible values for the angle theta between L and the z-axis a) for l = 1, b) l = 2.

a) First of all, we want to find the arccos(L_z /L). The ratio L_z/L can be simplified to m_l/(sqrt(l(l+1)) So we have theta = arccos(m_l/(sqrt(l(l+1)) ). Substituting 1 in for l, we have arccos(m_l/(sqrt(2)). Since m_l is related to L by ranging from L to L, we substitute -1, 0, 1 for m_l . The following angles are computed: 3pi/4, pi/2, pi/4. (radians)
b) Following the same procedure as in step a, we now have 5 values for m_l , -2, -1, 0, 1, 2. The equation is now theta = arccos(m_l/sqrt(6)). Solving for theta, we arrive at: 2.526, 1.991, pi/2, 1.150, 0.615 (radians).

Problem 26
What is the uncertainty in the mass of a muon (m = 105.7MeV/c^2), specified in eV/c^2, given its lifetime is 2.20*10^(-6)s?

We use the relation dEdt = h-bar. Solving for dE gives us 4.79*10^(-29)J. Dividing by e, we have 2.99*10^(-10)eV. Since dE = dmc^2, we have dm as 2.99*10^(-10)eV/c^2.

Problem 27
An electron and a 0.140kg baseball are each traveling 150m/s measured to an accuracy of 0.055%. Calculate and compare the uncertainty in position of each.

We use the relation dxdp = h-bar. Solving for dx in the electron, we arrive at 7.72*10^(-7)m. Since the accuracy is 0.055%, we divide by that and arrive at dx = 0.0014m.
Using the same relation, we solve for dx for the baseball, 5.02*10^(-36)m and divide by the accuracy again to arrive at 9.13*10^(-33)m.
*NOTE: we divide by accuracy because we would normally multiply dp by the accuracy.

Problem 28
Use the uncertainty principle to show that if an electron were present in the nucleus (r ~ 10^(-15)m) its kinetic energy ( use relativity ) would be hundreds of MeV.

Because we are given the uncertainty in position, we want to look for the uncertainty in momentum.
So, we want to look for this dp. dp can be solved for in the uncertainty equation dxdp = h-bar. dx is going to be the given r value. Next, we can use the relativity equation: E^2 = p^2c^2+m_0^2c^4. Solve for E, E = sqrt( p^2c^2+m_0^2c^4 ). E = sqrt( (h-bar/dx)^2c^2+m_0^2c^4 ). E is 3.16*10^(-11)J, 1.975*10^8eV. This is then 1.975*10^2MeV.

Problem 29
The uncertainty principle can be stated in terms of angular quantities as follows: dLdphi > h-bar. Here, L stands for angular momentum along a given axis and phi for the angular position measured in a plane perpendicular to that axis. a) Make a plausibility argument for this relation. b) Electrons in atoms have well-defined quantized values of angular momentum, with no uncertainty. What does this say about the uncertainty in angular position and the concept of electron orbits?

a) Start off by breaking down the terms into simpler quantities.
L = Iw. Now, relate linear quantities to angular quantities: w = v/r.
The moment of inertia for a hoop is I = mr^2.
We then have L = mr^2*v/r, L = mvr.
The length of the arc subtended by angle theta.. theta = l/r <==> Phi = x/r.
L*phi = mvr*x/r => Mvx
dMv*dx <==> dpdx.
b) If one has 0 uncertainty, then the other must have infinite uncertainty.

Problem 30
Show that there can be 18 electrons in a g sub-shell.

s, p, d, f, g. it is the fifth level. It then has 9 boxes. Each box can hold a maximum of 2 electrons, so there can only be 18 electrons.

Problem 31
The ionization energy of the outermost electron in boron is 8.26 eV. a) Use the Bohr model to estimate the effective charge seen by this electron. b) Estimate the average orbital radius.

a) First we determine what shell this electron is in. Boron has a configuration of 2s^2 2p^1 so it must be in the n = 2 level. Using equation E = -2pi^2*Z^2*e^4*mk^2/h^2 * 1/n^2, solving for Z^2 gives us 2.42, we then have Z as 1.557. OR better yet: E n = Z^2/n^2 * E 1 , E 1 = 13.6eV, E n = 8.26eV, n = 2. Z = 1.559.
b) The average orbital radius can be found by E n = kZe^2/r. First find the maximum orbital radius, r = 2.72*10^(-10)m. The average is then half of that, 1.36*10^(-10)m.