University of Rochester
Department of Physics and Astronomy
Physics123, Spring 2012
Recitation 11 - Solutions
Conceptual Questions:
•
When a wide spectrum of light passes through hydrogen gas at room temperature,
absorption lines are observed that correspond only to the Lyman series. Why do not
we observe the other series?
At room temperature, nearly all the atoms in hydrogen gas will be in the ground state.
When light passes through the gas, photons are absorbed, causing electrons to make
transitions to higher states and creating absorption lines. These lines correspond to the
Lyman series since that is the series of transitions involving the ground state or n = 1
level. Since there are virtually no atoms in higher energy states, photons corresponding to
transitions from n > 2 to higher states will not be absorbed.
•
How can you tell if there is oxygen near the surface of the Sun?
Look at a solar absorption spectrum, measured above the Earth’s atmosphere. If there are
dark (absorption) lines at the wavelengths corresponding to oxygen transitions, then there
is oxygen near the surface of the Sun.
Problem R11.1
Calculate the wavelength and the energy (in eV) of a photon whose frequency is
(a) 620 THz, (b) 3.10 GHz, and (c) 46.0 MHz.
State the classification of each on the electromagnetic spectrum.
d) At the atomic-scale, the electron volt and nanometer are well-suited units for energy
and distance, respectively. Demonstrate that when wavelength is expressed in nm and
energy in eV the general formula is valid
E=
1240 [eV ⋅ nm]
.
λ [nm]
Problem R11.1 – Solution
Energies are found using the formula for the single photon energy:
a) Visible light, blue. (See figure 31-12.)
1 c
3 × 10 8 ms-1
=
= 483.9 nm = 483.9 × 10 −9 m
ν 620 × 1012 s-1
1 eV
E = hν = (6.63 × 10 −34 J s)(
)(620 × 1012 s-1 ) = 2.57 eV
−19
1.60 × 10 J
b) Microwaves
c
3 × 10 8 ms-1
λ= =
= 9.68 cm = 0.0968 m
ν 3.10 × 10 9 s-1
1 eV
E = hν = (6.63 × 10 −34 J s)(
)(3.10 × 10 9 s-1 ) = 1.28 × 10 −5 eV
1.60 × 10 −19 J
c) Radio waves
c 3 × 10 8 ms-1
λ= =
= 6.52 m
ν 46 × 10 6 s-1
1 eV
E = hν = (6.63 × 10 −34 J s)(
)(46 × 10 6 s-1 ) = 1.91 × 10 −7 eV
−19
1.60 × 10 J
d)
c hc
1 eV
1 nm
1240
.
λ= =
= (6.63 × 10 −34 J s)(
) (3 × 10 8 ms)(
)≅
−19
−9
ν E
1.60 × 10 J
1 × 10 m
E
λ=
Problem R11.2
The spectrum of hydrogen (H, Z = 1) can be expressed simply in terms of the Rydberg
constant R, using formula 37-15:
1
⎛ 1
1 ⎞
= R⎜
−
,
2
⎝ (n ′ )
λ
(n)2 ⎟⎠
R=
Z 2 e4 m
8ε 20 h 3c
where λ is the wavelength of electromagnetic radiation emitted in vacuum, n1 and n2 are
integers such that n’ < n. The Rydberg constant is one of the most well-determined
physical constants, with a relative experimental uncertainty of less than 7 parts per trillion
(according to 2002 CODATA results).
a) Evaluate R for the hydrogen.
b) If the frequency of absorption from level n’ = 2 of the hydrogen is
νH = 799 191 727 402 000 Hz ,
what is the value of the final level n?
c) The same absorption experiment with deuterium (D, Z = 1) gives a frequency
νD = 799 409 184 967 000 Hz .
Such a discrepancy arises mainly from the evaluation of the Rydberg constant that must
be corrected by substituting m = me (the rest mass of the electron) with the reduced mass,
2 1
1
1
=
+
m me mn
where mn is the mass of the nucleus. Given νH and νD, determine mn/me.
d) What minimum resolving power is required for a grating to resolve those two H and D
absorptions?
e) Positronium (Ps) is a system consisting of an electron and its anti-particle, a positron,
bound together into an "exotic atom". The positron has the same mass of the electron and
opposite charge. The orbit of the two particles and the set of energy levels are similar to
that of the hydrogen atom. What is the energy of the ground state of the positronium?
Problem R11.2 - Solution
a) For hydrogen Z = 1.
⎛
⎞
⎛
⎞
1
1
1
Z 2e 4m 1
1
⎜
⎟
⎜
=R
− 2 = 2 3
− 2⎟ →
2
2
λ
⎜⎝ n ′
n ⎟⎠ 8ε 0 h c ⎜⎝ n ′
n ⎟⎠
( ) ()
( ) ()
()(
)(
4
2
)
1 1.602176 ×10 −19 C 9.109382 ×10 −31 kg
Z 2e 4m
R= 2 3 =
8ε 0 h c 8 8.854188 ×10 −12 C2 Nim 2 2 6.626069 ×10 −34 Jis 3 2.997925 ×108 m s
(
= 1.0974 ×107
4
C ikg
4
C
J 3s3 m s
2
N im 4
)(
)(
)
= 1.0974 ×107 m −1
b) n = 12 (it must be an integer).
c) mn/me = 3670.
d)
Δλ
Δν
=
= 2.72 × 10 −4 . Using Eq. 35-19, the minimum resolving power is 3676.
λ
ν
e) The only difference with hydrogen is the reduced mass. From Eq. 37-14a we have
E1 =
1
Z2
1
(−13.6 eV) 2 = − 13.6 eV = − 6.8 eV .
2
n
2
3 Problem R11.3
Take a free particle of mass m bouncing back and forth between two perfectly reflecting
walls, separated by distance l. Imagine that the two oppositely directed matter waves
associated with this particle interfere to create a standing wave with a node at each of the
walls. Show that the ground state (first harmonic) and the first excited state (second
2
h2
2 h
,
(2)
harmonic) have (non-relativistic) kinetic energies
, respectively.
8ml 2
8ml 2
Solution R11.3
For standing matter waves, there are nodes at the two walls. For the ground state (first
harmonic), the wavelength is twice the distance between the walls, or l = 12 λ (see Figure
15-26b). We use Eq. 37-7 to find the velocity and then the kinetic energy.
l = 12 λ → λ = 2 l ; p =
h
λ
2
=
h
p2
1 ⎛ h ⎞
h2
; K=
=
=
⎜ ⎟
2l
2m 2m ⎝ 2 l ⎠
8ml 2
For the second harmonic, the distance between the walls is a full wavelength, and so
l = λ.
2
h
p2
1 ⎛h⎞
h2
l =λ → p= = ; K =
=
⎜ ⎟ =
λ l
2m 2m ⎝ l ⎠
2ml 2
h
This is the same result of a free particle in a rigid box and it is not a coincidence. By
imposing nodes at each of the walls, we recreated the same boundary conditions.
4