Download 1 Data Provided: Formula sheet and physical constants

PHY332
Data Provided:
Formula sheet and physical constants
DEPARTMENT OF PHYSICS &
ASTRONOMY
Autumn Semester 2014-2015
ATOMIC AND LASER PHYSICS
2 hours
Answer question ONE (compulsory) and TWO other questions.
Please clearly indicate the question numbers on which you would like to be
examined on the front cover of your answer book. Cross through any work that you
do not wish to be examined.
The total number of marks for the exam paper is 50. Question ONE is marked out of
20. Questions 2-5 are marked out of 15. The breakdown on the right-hand side of the
paper is meant as a guide to the marks that can be obtained from each part.
PHY332
TURN OVER
1
PHY332
THIS QUESTION IS COMPULSORY.
1
(a)
Anti-hydrogen consists of a positron bound to an anti-proton. Positrons and antiprotons have the same mass but opposite charge to electrons and protons,
respectively. Calculate the wavelength of the n = 3 → 2 transition in anti-hydrogen
to three significant figures.
[2]
The upper and lower levels of a certain atomic transition have energies of –21112
and –33691 cm−1 respectively. Calculate the wavelength of the transition, and its
energy in eV units.
[2]
State, with reasons, which of the following transitions are forbidden for electric
dipole radiation:
2
P3/2 → 2S1/2 ;
4
I11/2 → 4H7/2 ;
3
F3 → 3 P2 ;
3
D1 → 1 P1 .
[2]
(d)
Estimate the energy of L shell absorption edge in lead (Z = 82).
[2]
(e)
Write down the electronic configuration of the ground state of arsenic (Z = 33) and
use Hund’s rules to deduce its values of the quantum numbers L, S and J.
[3]
What is the condition on the value of the spin quantum number S of the states
involved in the transition for a normal Zeeman effect to be observed? Explain why it
is only possible to observe the normal Zeeman effect in atoms with an even number
of electrons.
[3]
A laser transition in a certain atom has an Einstein A coefficient of 7.8×107 s–1. The
population of the upper level at time t = 0 is N0. What is the population of the upper
level at time t = 10 ns ?
[2]
The degeneracies of the upper and lower levels involved in the 694.3 nm transition
of the ruby laser 2 and 4 respectively. What is the effective temperature of the laser
levels when the population of the upper level in a working laser is equal to the
population of the lower level?
[4]
(b)
(c)
(f)
(g)
(h)
PHY332
CONTINUED
2
PHY332
2
(a)
(b)
(c)
(d)
He+ is a hydrogenic atom with an atomic number of 2. The gross-structure
spectrum would therefore fit to a Bohr-model formula of the type:
1
1
1
= 𝑅 ! − ! .
𝜆
𝑛! 𝑛!
Calculate the value of the effective Rydberg R to four significant figures, given that
𝑅! = 109 737 cm!! .
[2]
The radial part of the wave function of the 2p shell of He+ is of the form:
𝑅 𝑟 = 𝐶 𝑟 exp(− 𝑟 𝑎! ) ,
where a0 is the Bohr radius of hydrogen.
(i)
Show that the normalization constant C is equal to (2/√3) a0–5/2 .
(ii)
Find the value of the expectation value of the radius in units of a0.
You may use standard integrals on the formulae sheet without proof.
[6]
What value do you expect for the ionization energy of the He+ ion? Explain why it
is larger than the first ionization energy of neutral helium.
[3]
(i) Deduce the angular momentum quantum numbers L, S and J of the ground state
of neutral helium.
(ii) Write down the atomic states involved in the first three transitions of the
absorption spectrum of helium in order of increasing energy. In your answer you
should specify the electronic configurations and values of the quantum numbers L, S
and J of both the lower and upper levels involved in each transition.
3
[1]
[3]
PHY332
TURN OVER
PHY332
3
(a)
(b)
(c)
(d)
(e)
Discuss how evidence based on atomic radii and ionization potentials supports the
shell model of atoms.
[3]
The atomic number of chlorine is 17. Use this information to deduce whether there
is a non-zero electronic magnetic dipole moment in:
(i) chlorine atoms;
(ii) chlorine molecules (Cl2).
[2]
The alkali atom potassium has a ground state electronic configuration of [Ar] 4s.
Explain why the np → 4s transitions are split into doublets. (n is an integer ≥ 4.)
[3]
Explain the origin of hyperfine interactions in atoms and show that this leads to a
small energy shift for each atomic level of the form:
ΔEhyperfine = CHFS I⋅J.
In your answer you should state clearly the meanings of I and J.
[3]
The nuclear spin of the potassium 39 isotope (39K) is 3/2. How many hyperfine lines
would be observed for the following transitions in 39K atoms:
(i)
5p1/2 → 4s1/2 ;
(ii)
4p3/2 → 4s1/2 .
[4]
4
PHY332
CONTINUED
PHY332
4
(a)
(b)
(c)
In the context of broadening of atomic spectral lines, explain what is meant by:
(i)
natural broadening,
(ii)
Doppler broadening,
(iii)
pressure broadening.
[3]
The upper level of the 632.8 nm laser transition in neon has a radiative lifetime of
30 ns. The linewidth of the 632.8 nm transition in a low pressure neon lamp is found
to be 1.5 GHz. Discuss whether the transition is homogeneously or
inhomogeneously broadened, and deduce a value for the temperature of the atoms in
the lamp, given that the atomic weight of neon is 20.2.
[3]
The full width at half maximum (FWHM) of a narrow atomic transition of centre
wavelength λ0 is Δν. Show that the FWHM of the transition in wavelength units is
given by:
∆𝜆 =
(d)
(e)
!!
!
!
Δ𝜈 .
Find the value of Δλ for the 632.8 nm transition of neon in the discharge lamp
considered in part (b), expressing your answer in nm units.
[3]
Explain how population inversion is achieved for the 632.8 nm transition in a
helium-neon laser.
[3]
A helium-neon laser operating at 632.8 nm consists of a discharge tube of length
0.3 m with mirrors bonded to the ends. The reflectivities of the high reflector and
the output coupler are 99.8% and 98.5% respectively.
(i)
What is the gain coefficient of the laser medium?
(ii)
Calculate the separation of the longitudinal modes of the cavity.
[3]
Throughout this question, you may assume without proof that the Doppler linewidth of a transition of
wavelength λ for atoms with mass m is given by:
1/ 2
2 ⎛ 2 ln 2 kBT ⎞
Δν = ⎜
⎟
λ ⎝
m
⎠
PHY332
.
TURN OVER
5
PHY332
5 (a)
(b)
(c)
Explain how a laser beam tuned to near-resonance with an atomic transition can
be used to reduce the temperature of a beam of atoms moving with velocity v
towards the laser. Your answer should include a derivation of the optimal
frequency shift of the laser in terms of v relative to the centre frequency of the
atomic transition.
[3]
A beam of 133Cs atoms is emitted in the +x direction from an oven at a
temperature of 500 °C and is cooled by a laser beam directed in the –x
direction. The laser is tuned to near resonance with the 5s 2S1/2 → 5p 2P3/2
transition at 852 nm, which has a radiative lifetime of 30 ns.
Estimate:
(i)
the decelerating force applied to the atoms by the laser,
(ii)
the time taken to cool the atoms to their minimum temperature,
(iii)
the distance the atoms travel in this time,
(iv)
the final temperature achieved, stating any assumptions you make.
[4]
Explain what is meant by Bose–Einstein condensation. Why does it not occur
for fermionic particles? Show that the critical temperature Tc for Bose–Einstein
condensation is proportional to (N/V)2/3, where N is the total number of particles
in volume V.
[4]
(d) State, with reasons, which of the following systems would be expected to
undergo Bose–Einstein condensation, and where appropriate, calculate Tc for a
gas with 1.0 × 1018 particles m–3:
(i) helium 4: 4He,
(ii) potassium 40: 40K,
(iii) cesium 133: 133Cs,
(iv) dilithium molecules: (6Li)2.
You may assume without proof that the Bose-Einstein condensation
temperature is given by:
ℎ! 𝑁 !/!
𝑇! = 0.0839 ,
𝑚𝑘! 𝑉
where m is the particle mass. The atomic numbers of lithium, potassium and
cesium are 3, 19 and 55 respectively.
END OF QUESTION PAPER
6
[4]
PHYSICAL CONSTANTS & MATHEMATICAL FORMULAE
Physical Constants
electron charge
electron mass
proton mass
neutron mass
Planck’s constant
Dirac’s constant (~ = h/2π)
Boltzmann’s constant
speed of light in free space
permittivity of free space
permeability of free space
Avogadro’s constant
gas constant
ideal gas volume (STP)
gravitational constant
Rydberg constant
Rydberg energy of hydrogen
Bohr radius
Bohr magneton
fine structure constant
Wien displacement law constant
Stefan’s constant
radiation density constant
mass of the Sun
radius of the Sun
luminosity of the Sun
mass of the Earth
radius of the Earth
e = 1.60×10−19 C
me = 9.11×10−31 kg = 0.511 MeV c−2
mp = 1.673×10−27 kg = 938.3 MeV c−2
mn = 1.675×10−27 kg = 939.6 MeV c−2
h = 6.63×10−34 J s
~ = 1.05×10−34 J s
kB = 1.38×10−23 J K−1 = 8.62×10−5 eV K−1
c = 299 792 458 m s−1 ≈ 3.00×108 m s−1
ε0 = 8.85×10−12 F m−1
µ0 = 4π×10−7 H m−1
NA = 6.02×1023 mol−1
R = 8.314 J mol−1 K−1
V0 = 22.4 l mol−1
G = 6.67×10−11 N m2 kg−2
R∞ = 1.10×107 m−1
RH = 13.6 eV
a0 = 0.529×10−10 m
µB = 9.27×10−24 J T−1
α ≈ 1/137
b = 2.898×10−3 m K
σ = 5.67×10−8 W m−2 K−4
a = 7.55×10−16 J m−3 K−4
M = 1.99×1030 kg
R = 6.96×108 m
L = 3.85×1026 W
M⊕ = 6.0×1024 kg
R⊕ = 6.4×106 m
Conversion Factors
1 u (atomic mass unit) = 1.66×10−27 kg = 931.5 MeV c−2
1 astronomical unit = 1.50×1011 m
1 eV = 1.60×10−19 J
1 atmosphere = 1.01×105 Pa
1 Å (angstrom) = 10−10 m
1 g (gravity) = 9.81 m s−2
1 parsec = 3.08×1016 m
1 year = 3.16×107 s
Polar Coordinates
x = r cos θ
y = r sin θ
∂
1 ∂2
1 ∂
2
r
+ 2 2
∇ =
r ∂r
∂r
r ∂θ
dA = r dr dθ
Spherical Coordinates
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
dV = r2 sin θ dr dθ dφ
1
∂
1
∂2
1 ∂
∂
2
2 ∂
∇ = 2
r
+ 2
sin θ
+ 2 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
Calculus
f (x)
f 0 (x)
f (x)
f 0 (x)
xn
ex
nxn−1
ex
tan x
sin−1
ln x = loge x
1
x
sin x
cos x
cos x
− sin x
cosh x
sinh x
sinh x
cosh x
a
cos−1 xa
tan−1 xa
sinh−1 xa
cosh−1 xa
tanh−1 xa
cosec x
−cosec x cot x
uv
sec x
sec x tan x
u/v
sec2 x
x
√ 1
a2 −x2
− √a21−x2
a
a2 +x2
√ 1
x2 +a2
√ 1
x2 −a2
a
a2 −x2
0
0
u v + uv
u0 v−uv 0
v2
Definite Integrals
Z
∞
xn e−ax dx =
0
Z
+∞
n!
an+1
r
(n ≥ 0 and a > 0)
π
a
−∞
r
Z +∞
1 π
2 −ax2
xe
dx =
2 a3
−∞
Z b
b Z b du(x)
dv(x)
Integration by Parts:
u(x)
dx = u(x)v(x) −
v(x) dx
dx
dx
a
a
a
−ax2
e
dx =
Series Expansions
(x − a) 0
(x − a)2 00
(x − a)3 000
f (a) +
f (a) +
f (a) + · · ·
1!
2!
3!
n X
n n−k k
n
n!
n
Binomial expansion: (x + y) =
x y
and
=
(n − k)!k!
k
k
k=0
Taylor series: f (x) = f (a) +
(1 + x)n = 1 + nx +
ex = 1 + x +
n(n − 1) 2
x + ···
2!
x 2 x3
+ +· · · ,
2! 3!
sin x = x −
ln(1 + x) = loge (1 + x) = x −
Geometric series:
n
X
rk =
k=0
Stirling’s formula:
(|x| < 1)
x3 x5
+ −· · ·
3! 5!
x2 x3
+
− ···
2
3
and
cos x = 1 −
x2 x4
+ −· · ·
2! 4!
(|x| < 1)
1 − rn+1
1−r
loge N ! = N loge N − N
or
ln N ! = N ln N − N
Trigonometry
sin(a ± b) = sin a cos b ± cos a sin b
cos(a ± b) = cos a cos b ∓ sin a sin b
tan a ± tan b
1 ∓ tan a tan b
sin 2a = 2 sin a cos a
tan(a ± b) =
cos 2a = cos2 a − sin2 a = 2 cos2 a − 1 = 1 − 2 sin2 a
sin a + sin b = 2 sin 21 (a + b) cos 12 (a − b)
sin a − sin b = 2 cos 12 (a + b) sin 12 (a − b)
cos a + cos b = 2 cos 12 (a + b) cos 12 (a − b)
cos a − cos b = −2 sin 12 (a + b) sin 21 (a − b)
eiθ = cos θ + i sin θ
1 iθ
1 iθ
cos θ =
e + e−iθ
and
sin θ =
e − e−iθ
2
2i
1 θ
1 θ
cosh θ =
e + e−θ
and
sinh θ =
e − e−θ
2
2
sin a
sin b
sin c
Spherical geometry:
=
=
and cos a = cos b cos c+sin b sin c cos A
sin A
sin B
sin C
Vector Calculus
A · B = Ax Bx + Ay By + Az Bz = Aj Bj
A×B = (Ay Bz − Az By ) î + (Az Bx − Ax Bz ) ĵ + (Ax By − Ay Bx ) k̂ = ijk Aj Bk
A×(B×C) = (A · C)B − (A · B)C
A · (B×C) = B · (C×A) = C · (A×B)
grad φ = ∇φ = ∂ j φ =
∂φ
∂φ
∂φ
î +
ĵ +
k̂
∂x
∂y
∂z
∂Ax ∂Ay ∂Az
+
+
∂x
∂y
∂z
∂Ax ∂Az
∂Ay ∂Ax
∂Az ∂Ay
−
î +
−
ĵ +
−
k̂
curl A = ∇×A = ijk ∂ j Ak =
∂y
∂z
∂z
∂x
∂x
∂y
div A = ∇ · A = ∂ j Aj =
∇ · ∇φ = ∇2 φ =
∂ 2φ ∂ 2φ ∂ 2φ
+
+ 2
∂x2 ∂y 2
∂z
∇×(∇φ) = 0
and
∇ · (∇×A) = 0
∇×(∇×A) = ∇(∇ · A) − ∇2 A