A neutral Na atom has eleven electrons. Ten electrons are removed to
form a Na^{10+} ion. Calculate the frequency in Hz and the
wavelength in nm of the n = 2 to n = 1 transition of the Na^{10+} ion.

Solution:

- Concepts:

Hydrogenic atoms, scaling rules - Reasoning:

Na^{10+}is a single electron ion, the nuclear charge is Z = 11. - Details of the calculation:

E_{n}= -13.6 eV (μ'/μ)Z^{2/}n^{2}= -13.6 eV*121/n^{2}.

E_{2}- E_{1}= 121*(-3.4 + 13.6) eV = 1234.2 eV = hf = (4.136*10^{-15}eV s)*f. f = 2.98*10^{17}/s.

λ = c/f = 1.01 nm.

Assume a uniform electric field **E** = E**k** exists in some region of
space that contains no magnetic field. Let the magnetic vector potential
**A**(r,t) be zero. Write down an electrostatic potential Φ(r,t) for this field.

Define a gauge transformation, so that Φ'(r,t) = 0. What is the corresponding
vector potential** A**'?

Solution:

- Concepts:

Gauge transformations - Reasoning:

**A**and Φ are not unique. - Details of the calculation:

Φ(r,t) = -Ez is a possible electrostatic potential for this field.

Φ' = Φ - ∂λ/∂t = 0. λ = -Ezt.

**A**' =**∇**λ = -Et**k**.

We connect a real inductor to a source of alternating voltage with an
amplitude of 10 V and a frequency of 60 Hz. A current with an amplitude of
16 mA flows through this circuit. The amplitude of the current drops to 12
mA when we connect a resistor with a resistance of 500 Ω in series with the
inductor. Find the inductance L and the resistance R_{L} of the
inductor.

Solution:

- Concepts:

AC circuits - Reasoning:

A real inductor has some resistance. We model it as an ideal inductor L in series with a resistor R_{L}.

For any series RL circuit we have Z = R + iωL, |Z| = (R^{2}+ ω^{2}L^{2})^{½}, |I| = |V|/|Z|. - Details of the calculation:

1.6*10^{-2}A = 10/(R_{L}^{2}+ ω^{2}L^{2})^{½}, 1.2*10^{-2}A = 10/((R_{L}+ 500 Ω)^{2}+ ω^{2}L^{2})^{½}.

We have 2 equations for two unknowns.

In SI units:

R_{L}^{2}+ ω^{2}L^{2}= (10/1.6*10^{-2})^{2}= 625^{2}. ω^{2}L^{2}= 625^{2}- R_{L}^{2}.

(R_{L}+ 500)^{2}+ 625^{2}- R_{L}^{2}= 1000 R_{L}+ 500^{2}+ 625^{2}= 833^{2}.

R_{L}= 53.8 Ω.

ωL = 623 H/s, L = 1.65 H.

Consider a two-state system. The eigenbasis of the Hamiltonian H_{0}
is {|1>, |2>}, and in that basis the matrix of the H_{0} is

H_{0} = ħ

ω_{0} |
0 | ||

0 | ω_{0} |

.

Let H = H_{0} + W be the Hamiltonian of the system when a small
perturbation is applied, and let the matrix of W in the {|1>, |2>} basis
be

W = ħ

0 | ε_{1} |
||

ε_{2} |
0 |

.

Find the eigenvalues of H to first order using stationary perturbation theory, and find the corresponding zeroth order normalized eigenstates.

Solution:

- Concepts:

First order perturbation theory for degenerate states - Reasoning:

We have to diagonalize the matrix of W in the subspace of degenerate states. - Details of the calculation:

(a) det(W - λI) = 0 yields λ^{2}- ħ^{2}ε_{1}ε_{2}= 0.

λ_{±}= ± ħ(ε_{1}ε_{2})^{½}are the first-order corrections.

To first order the eigenvalues of H are

E_{+}= ħ(ω_{0}+ (ε_{1}ε_{2})^{½}) and E_{-}= ħ(ω_{0}- (ε_{1}ε_{2})^{½}).

The corresponding eigenstates are

|ψ_{+}> = N(|1> + (ε_{2}/ε_{2})^{½}|2>)

and

|ψ_{-}> = (|1> - (ε_{2}/ε_{2})^{½}|2>),

with N = (ε_{1}/(ε_{1}+ ε_{2}))^{½}.

A rectangular trough extends infinitely along the z direction, and has a cross section as shown in the figure.

All the faces are grounded, except for the
top one, which is held at a potential V(x) = V_{0}. Find the potential inside the
trough.

Solution:

- Concepts:

General solution to Laplace's equation, boundary conditions - Reasoning:

ρ = 0 inside the trough,**∇**^{2}V = 0 inside the trough. - Details of the calculation:

This is a two-dimensional problem. V does not depend on z.

In Cartesian coordinates**∇**^{2}V = ∂^{2}V/∂x^{2}+ ∂^{2}V/∂y^{2}= 0.

Separation of variables: V(x,y) = F_{1}(x)F_{2}(y).

(1/F_{1})∂^{2}F_{1}/∂x^{2}+ (1/F_{2})∂^{2}F_{2}/∂y^{2}= 0.

Since each term in the sum is a function of a different independent variable, each term must be equal to a constant and the constants have to sum to zero.

Let (1/F_{1})∂^{2}F_{1}/∂x^{2}= -C^{2}, ∂^{2}F_{1}/∂x^{2}+ C^{2}F_{1}= 0.

Then (1/F_{2})∂^{2}F_{2}/∂y^{2}= C^{2}, ∂^{2}F_{2}/∂y^{2}- C^{2}F_{2}= 0.

Solutions that can fulfill the boundary conditions on all faces with V = 0 require

F_{1}(x) = sin(Cx), C = nπ/b, F_{2}(y) = sinh(Cy), n = 1, 2, 3, ... .

The solution that satisfies all boundary conditions is

V(x,y) = ∑_{n=1}^{∞}D_{n}sin(nπx/b) sinh(nπy/b),

subject to the boundary condition V(x,a) = V_{0}.

V_{0}= ∑_{n=1}^{∞}D_{n}sin(nπx/b) sinh(nπa/b).

Property of the sine functions:

∫_{0}^{b}dx V_{0}sin(pπx/b) = 2V_{0}b/(pπ) if p is odd, 0 otherwise.

But ∫_{0}^{b}dx V_{0}sin(pπx/b) = ∑_{n=1}^{∞}D_{n}sinh(nπa/b) ∫_{0}^{b}dx sin(pπx/b) sin(nπx/b) = D_{p}sinh(pπa/b)(b/2).

Therefore D_{p }= 4V_{0}/(pπ sinh(pπa/b)), p = odd, 0 otherwise.

A very large sheet of electric charge densities +σ lies parallel to and a
small distance d_{1} in front of a very large grounded conducting plane.
A point charge +q is placed a distance d_{2} < d_{1} in front of
the plane, very far from the edges. The plane and the sheet are held in place.
Find the external force required to also keep the point charge at rest.

Solution:

- Concepts:

Gauss' law. method of images, - Reasoning:

This problem involves charge distributions in front of a conducting plane. - Details of the calculation:

Locate the grounded conducting plane at z = 0, the sheet at z = d_{1}, and put the point charge at on the z-axis at z = d_{2}.

The positively charged sheet induces a surface charge, which produces an electric field equal to that produced by an image sheet of charge densities -σ at a distance z = -d_{1}. The point charge +q also induces a surface charge distribution, which produces an electric field equal to that produced by an image charge -q on the z-axis at z = -d_{2}. The electric force acting on q is a superposition of the forces due to the sheet, the image sheet and the image charge.

The field due to an very large charged surface near the surface is found from Gauss' law.

On the z-axis, in the region 0 < z < d_{1}the field due to all charges except q itself is=

E**E**(+σ) +**E**(-σ) +**E**(q) =**k**[-σ/(2ε_{0}) - σ/(2ε_{0}) - q/(4πε_{0}(z + d_{2})^{2})].

Electric force on q:**F**_{el}= q**E**= -**k**[ qσ/ε_{0}+ q^{2}/(16πε_{0}d_{2}^{2})].

External force required to also keep the point charge at rest:**F**_{ext}= -**F**_{el}.