Problem 1:
A hydrogen atom is known to be in a state characterized the
quantum numbers n = 4, l
= 2.
(a)
Give the allowed values of j.
(b)
For each of the allowed values of j, calculate the square of the
magnitude of the total angular momentum.
Solution:
 Concepts:
Addition of angular momentum
 Reasoning:
We are supposed to add the
orbital and spin angular momentum of the electron in the hydrogen atom.
 Details of the calculation:
(a)
l = 2, s = ½, j = 5/2, 3/2.
(b)
J^{2}jm> = j(j + 1)ħ^{2}jm>.
For j = 5/2, j(j + 1)ħ^{2}
= (35/4)ħ^{2}.
For j = 3/2, j(j + 1)ħ^{2}
= (15/4)ħ^{2}.
Problem 2:
An electromagnetic wave with frequency 65.0 Hz travels
in an insulating magnetic material that has dielectric constant κ_{e}
= 3.64 and
relative permeability κ_{m} = 5.18 at
this frequency. The electric field has amplitude 7.20*10^{−3 }V/m.
(a) What is the speed of propagation of the wave?
(b) What is the wavelength of the wave?
(c) What is the amplitude of the magnetic field?
Solution:

Concepts:
EM waves

Reasoning:
We are supposed to recall some basic properties of EM waves.

Details of the calculation:
(a) v = c/(κ_{e}κ_{m})^{½} = 6.91*10^{7} m/s.
(b) λ = v/f = 1.06*10^{6} m.
(c) B_{max} = E_{max}/v = 1.04*10^{10} T.
Problem 3:
Consider an excited Nitrogen atom with a an electronic configuration
(1s)^{2}, (2s)^{2}, (2p)^{2}, (3s).
Find the spectroscopic terms ^{2S+1}L_{J} that characterize the
system.
Solution:
Problem 4:
A particle with mass m is confined to a 3D box.
V(x, y, z) = 0 if 0 ≤ x ≤ a AND 0 ≤ y ≤ b AND 0 ≤ y ≤
c,
V(x, y, z) = ∞ otherwise.
(a)
Write down expressions for the energy eigenfunctions and eigenvalues for this
particle.
(b) Assume that the particle is in the ground state of the system.
At time t = 0,
all dimensions of the box suddenly double, i.e. a > 2a, b > 2b, c > 2c. A measurement of
the energy of the particle is made just after the dimensions increase. (Assume that all
this happens so quickly so the spatial wave function of the particle does not
change). What is the probability that the energy measurement yield a value
EXACTLY the same as the energy of the ground state of original system?
Solution:
 Concepts:
The particle in a 3dimensional box, separation of variables
 Reasoning:
We are asked to find the eigenfunctions and eigenvalues of the Hamiltonian
of a particle in a 3D box.
 Details of the calculation:
(a)
Φ_{nx,ny,nz}(x,y,z) = (8/(abc))^{½}sin(n_{x}πx/a) sin(n_{y}πy/b) sin(n_{z}πz/c).
E_{nx,ny,nz} = π^{2}ħ^{2}[n_{x}^{2}/(2ma^{2})
+ n_{y}^{2}/(2mb^{2}) + n_{z}^{2}/(2mc^{2})].
(b) For the original system
E_{ground} = E_{1,1,1} = π^{2}ħ^{2}[1/(2ma^{2})
+ 1/(2mb^{2}) + 1/(2mc^{2})].
For the new system we have
Φ^{'}_{nx,ny,nz}(x,y,z) = (1/(abc))^{½}sin(n_{x}πx/(2a)) sin(n_{y}πy/(2b)) sin(n_{z}πz/(2c)).
E^{'}_{nx,ny,nz} = π^{2}ħ^{2}[n_{x}^{2}/(8ma^{2})
+ n_{y}^{2}/(8ma^{2}) + n_{z}^{2}/(8ma^{2})].
An energy measurement yield a value EXACTLY the same as the energy of the ground
state of original system if the system is in the found in the state
Φ^{'}_{2,2,2}(x,y,z).
The probability of finding the system in that state is <Φ_{1,1,1}Φ^{'}_{2,2,2}>^{2}.
<Φ_{1,1,1}Φ^{'}_{2,2,2}>^{2} = (8/(abc)^{2})∫_{0}^{a}sin^{2}(πx/a)dx
∫_{0}^{b}sin^{2}(πy/b)dy ∫_{0}^{c}sin^{2}(πz/a)dz^{2}
= 1/8.
The probability that the energy measurement yield a value EXACTLY the same as
the energy of the ground state of original system is 1/8 = 12.5%.
Problem 5:
A nonconducting sphere of radius R = 0.1 m, mass M = 10 kg and uniform mass
density carries a surface charge density σ = σ_{0}cosθ, with σ_{0}
= 10 microCoulomb.
(a) Find the dipole moment p_{0} of
the sphere.
(b) Assume that at t = 0 the sphere receives an angular
impulse and starts rotating about the xaxis with angular velocity ω_{0 }
= 1000/s. Calculate the power radiated by the sphere when it rotates with
angular frequency ω.
(c) Estimate the time it takes for the rotation
rate of the sphere to decrease by a factor of 2.
Solution:
 Concepts:
Dipole moment, the Larmor formula
 Reasoning:
The y and z components of the dipole moment are
oscillating with angular frequency ω. The rotating electric dipole will
radiate energy at a rate P_{rad} = <(d^{2}p/dt^{2})^{2}>/(6πε_{0}c^{3}).
 Details of the calculation:
(a) Dipole moment: p = ∫σ(r)r
dA. Symmetry dictates that p = p k.
p = k ∫_{0}^{π}σ_{0}cosθ Rcosθ 2πR^{2}sinθdθ
= k σ_{0}(4π/3)R^{3} = k p_{0}.
p_{0}
= 4.19*10^{8} Cm.
(b) p(t) = p_{0}cos(ωt) k
+ p_{0}sin(ωt) j.
d^{2}p/dt^{2} =
ω^{2}p, <(d^{2}p/dt^{2})^{2}> = ω^{4}p_{0}^{2}.^{
}P_{rad} = <(d^{2}p/dt^{2})^{2}>/(6πε_{0}c^{3})
= ω^{4}p_{0}^{2}/(6πε_{0}c^{3}).
<P> = ω^{4}*3.9*10^{31} W s^{4}.
For ω_{
}= ω_{0 }= 1000/s we have <P> = 3.9*10^{19} W.
(c)
<P> = dE/dt.
E = ½Iω^{2} = (1/5)MR^{2}ω^{2}.
dE/dt = (2/5)MR^{2}ω
dω/dt = ω^{4}p_{0}^{2}/(6πε_{0}c^{3}).
5p_{0}^{2}dt/(12 MR^{2}πε_{0}c^{3})
= dω/ω^{3}.
[5p_{0}^{2}/(12 MR^{2}πε_{0}c^{3})]
∫_{0}^{T} dt = ∫_{ω0} ^{ω0/2} dω/ω^{3}.^{
}[5p_{0}^{2}/(12 MR^{2}πε_{0}c^{3})]
T = ½((ω_{0}/2)^{2}  ω_{0}^{2}).
T ≈ 10^{23} s.
Problem 6:
Starting with the transformation of the electromagnetic fields under a
Lorentz transformation show that
(a) if E is normal to B in an inertial frame, it must be true in
all other inertial frames, and
(b) if E > cB in an inertial frame, it must be true in all other inertial
frames.
Solution:
 Concepts:
Vector products
 Reasoning:
Lorentz transformation of the electromagnetic fields:
E'_{} = E_{},
B'_{} = B_{},
E'_{⊥} = γ(E + v×B)_{⊥},
B'_{⊥} = γ(B  (v/c^{2})×E)_{⊥}.
We can show that (E∙B)^{2} and
E^{2 }
c^{2}B^{2} are invariant under a Lorentz
transformation.
 Details of the calculation:
(a) We can choose a Cartesian coordinate system and evaluate the
products using Cartesian coordinate.
Consider to reference frames, K, and K'. Let the direction of the relative
velocity of K' with respect to K be the zdirection, v = vk.
Then E_{} = E_{z} k and E_{⊥}
= E_{x} i + E_{y} j,
B_{} =
B_{z} k and B_{⊥} = Bxi +
B_{y} j,
E∙B = E_{x}B_{x} + E_{y}B_{y}
+ E_{z}B_{z}.
E'∙B' = E'_{}∙B'_{} +
E'_{⊥}∙B'_{⊥}.
E'_{}∙B'_{} = E'_{z}B'_{z}
= E_{z}B_{z}.
E'_{⊥} = E'_{x} i + E'_{y}
j.
B'_{⊥} = B'_{x} i + B'_{y}
j.
E'_{x}= γ(E_{x}  vB_{y}), E'_{y}= γ(E_{y}
+ vB_{x}), B'_{x}= γ(B_{x} + (v/c^{2})E_{y}),
B'_{y}= γ(B_{y}  (v/c^{2})E_{x}).
E'_{x}B'_{x} = γ^{2}(E_{x}B_{x} +
(v/c^{2})E_{x}E_{y}  vB_{y}B_{x} 
(v^{2}/c^{2})B_{y}E_{y}).
E'_{y}B'_{y} = γ^{2}(E_{y}B_{y} 
(v/c^{2})E_{y}E_{x} + vB_{x}B_{y} 
(v^{2}/c^{2})B_{x}E_{x}).
E'_{x}B'_{x} + E'_{y}B'_{y} = γ^{2}(1
 v^{2}/c^{2})(E_{x}B_{x} + E_{y}B_{y})
= E_{x}B_{x} + E_{y}B_{y}.
(E∙B)^{2} is invariant under a Lorentz transformation.
If E∙B is zero in one frame, it is zero in every frame.
(b) E^{2 } c^{2}B^{2} = E_{x}E_{x}
+ E_{y}E_{y} + E_{z}E_{z}  c^{2}(B_{x}B_{x}
+ B_{y}B_{y} + B_{z}B_{z}).
E'_{z}E'_{z}  c^{2}B'_{z}B'_{z} = E_{z}E_{z}
 c^{2}B_{z}B_{z}.
E'_{x}E'_{x} = γ^{2}(E_{x}^{2}  2vE_{x}B_{y}
+ v^{2}B_{y}^{2}).
E'_{y}E'_{y} = γ^{2}(E_{y}^{2} + 2vE_{y}B_{x}
+ v^{2}B_{x}^{2}).
B'_{x}B'_{x} = γ^{2}(B_{x}^{2} +
2(v/c^{2})E_{y}B_{x} + (v^{2}/c^{4})E_{y}^{2}).
B'_{y}B'_{y} = γ^{2}(B_{y}^{2} 
2(v/c^{2})E_{x}B_{y} + (v^{2}/c^{4})E_{x}^{2}).
E'_{x}E'_{x} + E_{y}'E'_{y}  c^{2}(B'_{x}B'_{x}
+ B'_{y}B'_{y})
= γ^{2}(E_{x}^{2} + v^{2}B_{y}^{2}
+ E_{y}^{2} + v^{2}B_{x}^{2}  c^{2}B_{x}^{2}
 (v^{2}/c^{2})E_{y}^{2}  c^{2}B_{y}^{2}
 (v^{2}/c^{2})E_{x}^{2})
= γ^{2}(1  v^{2}/c^{2})(E_{x}^{2} +
E_{y}^{2})  γ^{2}(c^{2}  v^{2})(B_{x}^{2}
+ B_{y}^{2}) = E_{x}E_{x} + E_{y}E_{y}
 c^{2}(B_{x}B_{x} + B_{y}B_{y}).
E^{2 } c^{2}B^{2} is invariant under
a Lorentz transformation.
If E > cB in one inertial frame then E > cB in every inertial frame.