Boltzmann's constant k= 1.38066E-23 J/K
k= 0.00008617 eV/K
Atomic mass unit u= 1.66E-27 kg
umev= 931.5 MeV/c^2
Wien constant 2.90E-03
Planck's constant h= 4.14E-15 eV s
h= 6.63E-34 J s
h/ 2 pi hbar 1.05E-34 J s
h/ 2 pi hbev 6.59E-16 eV s
Speed of light c= 3.00E+08 m/s
hc= 1.24E+03 eV nm
hbarc= 1.97E+02 eV nm
Electron charge e= 1.60E-19 C
electron volt eV= 1.60E-19 joule
electron mass mel 9.11E-31 kg
mel 5.11E-01 MeV/c^2
proton mass mp 1.67E-27 kg
mpev 9.38E+02 MeV/c^2
neutron mass mn 1.67E-27 kg
mnev 9.40E+02 MeV/c^2
Stefan's constant= sigma= 5.67E-08 watt/m^2K^4
Nuclear distance unit fermi 1.00E-15 m (a femtometer)
cross section unit barn= 1.00E-28 m^2
barn= 100 fm^2
common combo for scattering ke^2 1.44 MeV fm
First Bohr orbit radius a0 a0 0.0529 nm
Bohr magnetion mub 5.79E-05 eV/T
muba 9.27E-24 A m^2 A=ampere
#2 a.Use radial probability to estimate energy of ground state.
Magnitude of average kinetic energy is equal to that of average potential energy.
Average PE equal to potential energy at the expectation value for radius, ke^2/<r>
The expectation value for the radius is on the order of 0.1nm
Average KE = 1.44 eV nm/0.1 nm = 14.4 eV
2.b. From the probability curve, the distribution is mostly contained
within +/- a Bohr radius, so uncertainty in
the neighborhood of 1 Bohr radius.
2c. The uncertainty in momentum is on the order of the momentum
= sqrt(2*m*<E>) = 3836.248167 eV/c
Checking for consistency with the uncertainty principle
delta x * delta p = 3836 eV/c *.0529 nm = 202.9244 eV nm/c 1.030073096 hbar
so these estimates are consistent with the uncertainty principle.
#3 Probability that electron in hydrogen is closer than 1 Bohr radius.
The value of the probability for any range of r is given by
P = 4[exp(-2*r/a0)*(-a0*r^2/2 - a0^2*r/2 -a0^3/4)]/a0^3 evaluated between the radius limits.
For limits 0 to a0, the terms in parentheses above reduce to -5/4 at a0 and
to -1/4 at r=0. So the probability is exp(-2)*(-5) +1 = 0.323323584
#4 Probability that electron is between 1 and 4 Bohr radii if it is
in ground state.
The probability stated in #3 evaluated between the limits a0 and 4*a0 gives
=4*(exp(-8)*(-16/2-4/2 -1/4) -exp(-2)*(-5))= 0.662922448
#9 Probability that 2s and 2p electrons will be found inside the nucleus.
Since the nuclear radius is 3 orders of magnitude less that the Bohr radius
the exponential terms in the wavefunctions may be set equal to 1
The 2s wavefunction integral can then be approximated by the integral of
r^2*(2 - r/a0)^2/(2*a0^3) or r^2*(1 + r^2/(4*a0^2) - r/a0)/(2*a0^3)
The integral between zero and the nuclear radius R is
P = R^3/(6*a0^3) + R^5/(40*a0^5) - R^4/(8*a0^4)
Taking the radius R=1 fermi, this becomes
P = (1E-15 m)^3/(6*(.529E-10 m))^3 = 1.12585E-15
This compares to about 1E-14 for the 1s state, so a 1s electron is 10x more likely to
be found inside the nucleus.
b. For the 2p state, the integrand with the exponential set equal to 1 is
The integral between zero and the nuclear radius R is
R^5/(120*a0^5) = P = (1E-15 m)^5/(120*(.529E-10 m)^5 = 2.01159E-26
This is 12 orders of magnitude less than the probability for the 1s electron.
#11 Values for L and Lz for 3p and 3d states.
a. 3p state
l=1, L = sqrt(l*(l+1))*hbar = sqrt(3)*hbar
ml=-1,0,1 and Lz = ml*hbar
b. 3d state
l=2, L = sqrt(l*(l+1))*hbar = sqrt(6)*hbar
ml=-2,-1,0,1,2 and Lz = ml*hbar
#12 Quantum numbers for n=5 states
#18 Beam splitting in Stern-Gerlach experiment
a. The force on the atoms is F = -muz * dB/dz
The z component of the magnetic moment is muz = -e*hbar/(2*m) = -1 Bohr magneton
The speed of the atoms is v = sqrt(2*KE/M), where M is the mass of the silver atom.
The time that the atoms experience the force is t = L/v = L/sqrt(2*KE/M)
The lateral displacement is given by x = .5*a*t^2 = mub*dB/dz*M*L^2/(2*M*2*KE)
The separation is twice this displacement = (mub*L^2/(2*KE))*dB/dz where mub=Bohr magneton
b. Calculation of field gradient necessary to give a mm of separation of the beams on the film
The KE=2kT for particles escaping through a hole from a hot oven (not 3kT/2 !).
The field gradient is given by dB/dz = 2*KE*d/(mub*L^2) = 4kTd/(mub*L^2)
dB/dz= (4*k*1000*.001)/(mub*.05^2) = 2.38E+03 T/m
#21 Quantum numbers related to total angular momentum
a. The possible values of mj are -j, -j+1, … j which give a total of 2j + 1 values
b. Quantum numbers of the n=2 states in terms of hydrogen in terms of n, l, ml, and ms
Quantum numbers of the n=2 states in terms of n,l,j,mj
c. If mj=3/2, what are the possible values of l, ml, and ms?
This is the stretched case where l=1 and ml=1 and ms=1/2
d. If mj=1/2, what are the possible values of l, ml, and ms?
There are several ways to get mj=1/2
#27 Zeeman splitting for hydrogen 3->2 transition
The photon energy = 13.6 eV*(1/4 - 1/9) = 1.888888889 eV
You get 7 d states and 5 p states
The width of the set of d states is then 6*mub*B and the width of the p states is 4*mub*B
The rms width of the transition is then sqrt((delta Ed)^2 + (delta Ep)^2)= sqrt(36+16)*mub*B
delta E/E = 1E-4 = sqrt(36+16)*mub*B/1.89
This corresponds to a field B = 4.53E-01 Tesla
#29 Modeling electron spin with a spinning sphere
Angular momentum L = I*omega = hbar/2
For a sphere, I = 2*M*r^2/5
omega = L/I = (hbar/2)/(2*M*r^2/5) = 1.45E+32 1/s
Not a good model since electron size (spatial extent, wavelength, uncertainty in position)
depends upon its momentum
#31 2p state of hydrogen
a. most probable radius for 2p electron.
Radial part of probability = 4*pi*r^2*wavefunction^2
Take derivative to find most probable radius
d/dr(r^2*(r/a0)^2*exp(-r/a0)) = (4r^3/a0^2)*exp(-r/a0) - (r^4/a0^3)*exp(-r/a0) =0
This gives a most probable r = 4*a0
b. Probability of being within 10% of this most probable value
Integrate the probability distribution between 3.6 and 4.4 a0.
Can approximate by taking peak probability and multiplying by the interval 0.8*a0
This shakes down to (1/(24*a0^3))*(0.8*a0)*(4*a0)^4/a0^2)*exp(-4) = 0.156293452