Download Page 1 PHYSICS 4100 Modern Physics Second Examination

PHYSICS 4100 Modern Physics
Second Examination (Chapters 3–5)
Answer All Questions
February 24, 2012
Name SOLUTIONS
1. (a) In a Millikan oil drop experiment, a drop that has a charge +e is observed to be moving vertically upward at terminal
velocity. The drop suddenly loses another electron and soon thereafter attains a new terminal velocity. Show that though
the net charge has doubled, the terminal velocity has not. (b) In a Thomson e/m experiment, a voltage is applied between
parallel plates 0.05 m long. Electrons are observed to be deflected by an angle of 120 milliradians after traveling the
length of the plates. When a magnetic field of 10–4 T is applied, there is no deflection. Calculate the speed of the electrons.
[Hint: Obtain two equations involving the electric field and the speed, and solve for the speed. Use e/m = 1.76 × 1011
C/kg.] (20 points)
(a) The free-body diagram representing the situation is shown to the right. At the low speeds of the
drops in the Millikan experiment, the drag force is proportional to the velocity. At terminal velocity,
the net force is zero, so qE = mg+buT, where uT is the terminal speed. (The velocity is upward.) Thus,
uT = (qE – mg)/b. Note that the terminal speed is linearly related to q, but not proportional to q.
Thus, if q is doubled, the terminal speed (and hence the terminal velocity) will not be doubled. (5)
qE
mg
Fdrag= bu

(b) Let us assume the +x direction is in direction of theinitial velocity u of the electrons and the +y
direction is in the direction opposite the electric field E . The y-component of the velocity of an electron exiting the
region between the plates is given by
eE L
uy = ayt =
, where L is the length of the plates. The deflection angle is given by
m u
u y eEL
tan θ =
=
. Because θ is small, tan θ ≈ θ . Hence, to good approximation,
u
mu 2
eEL
θ=
. ----- (1)
mu 2
Now, with an applied magnetic field, the deflection is zero when the Lorentz force is zero, i.e.,
euB = eE, or, E = uB. ------ (2)
Substituting for E in Eq. (1) gives θ =
eBL
mu
. Thus,
⎛ (10-4 T)(0.05 m) ⎞
u=
= (1.76 × 10 C/kg) ⎜
= 7.3 × 106 m/s. (15)
⎟
m θ
⎝ 0.120 rad ⎠
e BL
11
2. (a) Light of wavelength 200 nm is incident on a metal surface. The cutoff wavelength for the metal is 425 nm. (i)
Calculate the stopping voltage. (ii) If the intensity of the light is 2.0 W/m2, find the average number of photons per unit
area per unit time striking the surface. (b) A beam of photons strikes a target and the maximum kinetic energy of the
emerging Compton electrons is observed to be 5.0 keV. Determine the wavelength of the scattered photons. (20 points)
(a) (i) The work function φ = hfcutoff =
K max = hf − φ =
hc
λ
hc
1240 eV ⋅ nm
=
= 2.92 eV. From the Einstein equation,
λ cutoff
425 nm
− φ . But, K max = eVs. Hence, eVs =
hc
λ
− φ , or, Vs =
hc 1
e λ
−
φ
e
.
⎛ 1 ⎞ − 2.92 V = 3.28 V.
⎝ 200 nm⎠
Thus, Vs = 1240 V ⋅ nm
(ii) No. of photons per unit area per unit time = Intensity/energy per photon. Thus,
I
2.0 W/m 2
2.0 W/m 2
n=
=
= 1240 eV⋅nm 1.6 ×10 J = 2.0 × 1018 photons/m 2 / s. (10)
E photon
hc λ
( 200 nm ) 1 eV
(
-19
)
(b) The electron has maximum kinetic energy when the photon is scattered at 180°. Conservation of momentum gives
pi = pe − p f , where the subscripts e and i denote the incident and scattered photon,
respectively. Thus,
pi + p f = pe . ---- (1)
Conservation of energy gives
pi c − p f c = K e,max . ---- (2)
Multiplying Eq. (1) by c and subtracting Eq. (2) gives
2p f c = pe c − K e,max , i.e., p f c = ( pe c − K e,max ) / 2. ----- (3)
Now, pe c =
(
p c= (
2
pf c =
f
1
2
1
Ee2 − m2 c 4 =
( K e,max + mc 2 ) 2 − m2 c 4 =
2
K e,max
+ 2K e,max mc 2 . Thus, Eq. (3) becomes
)
2
K e,max
+ 2K e,max mc 2 − K e,max . Substituting values yields
)
(5.0 × 103 eV) 2 + 2(5.0 × 103 eV)(5.11 × 105 eV) − (5.0 × 103 eV) = 3.33 × 10 4 eV.
Finally, λ f =
h
pf
=
hc
pf c
=
1240 eV ⋅ nm
3.33 × 10 eV
4
= 0.0372 nm.
(10)
[This problem may also be solved using the expressions involving the wavelengths of the incident and scattered photons,
as done in a homework problem. A quadratic equation has to be solved.]
Modern Physics, Examination 2
Shand (2/23/12)
Page 2 of 3
3. (a) Consider a doubly ionized lithium atom Li++ in the ground state. The absorption of a photon causes the remaining
electron to be ejected from the Li++ ion, with the liberated electron having a kinetic energy of 15.0 eV. Determine the
wavelength of the photon using the Bohr theory. [Hint: What is the value of n when the electron escapes?] (b) (i) Use the
Bohr theory to estimate the wavelength of a copper (Z = 29) Kα x-ray photon (n = 2 → n = 1 transition). (ii) The actual
value of the wavelength of a copper Kα x-ray photon is 0.154 nm. Is your calculated value greater or smaller than the
actual value? Explain why. (20 points)
(a) By energy conservation, we have
E photon = ΔEion + K electron . ----- (1)
For doubly-ionized lithium (Z = 3), the change in energy (excitation energy) of the ion is given by
⎛ 1
ΔE = Z 2 (13.6 eV) ⎜
2
⎝ ni
−
1⎞
⎛1 1⎞
⎟ = (9)(13.6 eV) ⎜ 2 − 2 ⎟ = 122.4 eV.
⎝1 ∞ ⎠
nf ⎠
2
Substituting values into Eq. (1) gives E photon = 122.4 eV + 15.0 eV = 137.4 eV.
λ photon =
hc
E photon
=
1240 eV ⋅ nm
137.4 eV
= 9.02 nm.
(10)
(b) (i) For one-electron atom with an atomic number Z, the energy of a photon emitted when the atom makes a transition
from level ni to level nf is given by
E ph =
hc
λ
⎛ 1
= Z 2 ER ⎜
⎝ nf
2
−
1⎞
ni2 ⎟⎠
. Thus, the wavelength of the photon can be calculated from
−1
⎡ Z 2 ER ⎛ 1 1 ⎞ ⎤
λ=⎢
⎜ 2 − 2 ⎟ ⎥ . Substituting values yields
⎣ hc ⎝ n f ni ⎠ ⎦
−1
⎡ (29)2 (13.6 eV) ⎛ 1 1 ⎞ ⎤
λ=⎢
− 2 ⎥ = 0.144 nm.
2
⎣ 1240 eV ⋅ nm ⎝ 1 2 ⎠ ⎦
(7)
(ii) The calculated value is smaller than the actual value. The reason is an electron in the K shell (n = 1) is shielded from
the full nuclear charge by the other electron in the K shell. Thus, the effective nuclear charge “seen” by a K electron is
somewhat smaller than Ze. If we account for this effect by replacing Z in the above formulas with Z − δ , where δ is a
small positive number that represents the shielding effect, we see that a larger value of wavelength would be predicted, in
better agreement with the accepted value. (3)
Modern Physics, Examination 2
Shand (2/23/12)
Page 3 of 3