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Homework week 1 Chemistry 455A
Wednesday: Q12.3, Q12.4, P12.9, P12.17,
Friday: P12.20, P12.
Q 1.3/12.3) Why does the analysis of the photoelectric effect based on classical physics
predict that the kinetic energy of electrons will increase with increasing light intensity?
In the classical theory, the energy of the light is the light intensity. So to increase the
light’s energy just focus the light down on a smaller surface. Then the electron that is
ejected must move faster because it was hit with higher energy light.
Classical physicists thought that the greater energy would manifest as a higher speed in
the ejected electron. In fact, the light energy appears as a larger number of electrons
leaving the surface. The kinetic energy of the electrons, surprisingly, is determined by
the photon frequency and the work function of the surface.
Q1.4/12.4) What did Einstein postulate to explain that the kinetic energy of the emitted
electrons in the photoelectric effect depends on the frequency? How does this postulate
differ from the predictions of classical physics?
Einstein postulated that the energy of light depends on the frequency, whereas in
classical theory, the energy depends only on the intensity and is independent of
the frequency. He further postulated that the total energy of light would be a
multiple of hv , and thereby the photon was conceived.
P1.9/P12.9) A newly developed substance that emits 225 W of photons with a
wavelength of 225 nm is mounted in a small rocket such that all of the radiation is
released in the same direction. Because momentum is conserved, the rocket will be
accelerated in the opposite direction. If the total mass of the rocket is 5.25 kg, how fast
will it be traveling at the end of 365 days in the absence of frictional forces?
There is a short way to do this problem: One that shows that the velocity of the rocket
does not depend on the wavelength of the light. The total energy of the light field divided
by the speed of light is the total momentum, which can be equated to the rocket’s
momentum:
( )
total total total rocket photon
E E E
mv pn p p E cp c
The light leaves with energy, it is not all consumed by the rocket. So you cannot just
equate the energy of the rocket to the energy of the photons that left. But you can
equation the momentum transfer. From Einstein’s famous formula for light:
photon and
total photon
total photon photon
h E cp p
E nE
dE dn Power E n E dt dt
The number of photons per second is given by
1 1 20 1 34 8 1
9
1 J s PowerW W 225J s 2.547 10 s 6.626 10 J s 2.998 10 m s
225 10 m
total
photon
Power n E hc
− − − − −
−
×
′ = = = = ×
× × ×
×
The momentum of one photon is
34 27 1 9
6.626 10 J s 2.945 10 kg m s 225 10 m
h p
− − − −
×
= = = ×
×
The total number of photons is given from the energy of one photon and total energy:
1 9
27 1 19
28
9
8
86400 s 225 J s 365days =7 10 day
2.945 10 kg m s 8.8 10
Total
photon
Total
photon
Total
Total
E Power Time Joules
E cp Joules
E
n photons E
E
np kgm s c
E v m s c m
−
− − −
= ⋅ = ⋅ × ⋅
Now equate the momentum of the light beam, with the momentum of the rocket:
( )
28 27 0.8 10 2.945 10
mv (^) rocket np light
v m s m s
For a second way to compute the velocity using the forces:
The force is given by the rate of change of momentum.
( ) (^27 1 20 1 7 ) 2.945 10 kg m s 2.547 10 s 7.501 10 kg ms
d np F n p dt
− − − − − = = ′ = × × × = ×
The final speed is given by
7 2 1 0
7.501 10 kg m s 86400s v v 365days = 4.51m s 5.25 kg day
F
at t m
− − × − = + = = × ×
The total energy of the photon beam is
1 86400s 9 225J s 365days =7 10 day
E Total Power Time Joules
− = ⋅ = ⋅ × ⋅
P12.21 ) The work function of platinum is 5.65 eV. What is the minimum frequency of
light required to observe the photoelectric effect on Pt? If light with a 150-nm wavelength
is absorbed by the surface, what is the velocity of the emitted electrons?
a) For electrons to be emitted, the photon energy must be greater than the work function
of the surface.
19 19
19 15 1 34
1.602×10 J
5.65 eV 9.05 10 J eV
9.05 10 J 1.4 10 s 6.626 10 J s
E h
E
h
− −
− − −
×
×
This photon is in the UV; no wonder Pt is inert.
b) The outgoing electron must first surmount the barrier arising from the work function,
so not all the photon energy is converted to kinetic energy.
2
34 8 1 19 19 9
19 5 1 31
1 2 v
6.626 10 J s 2.998 10 m s 9.05 10 J 4.19 10 J 150 10 m
2 2 4.19 10 J
v = 9.59 10 m s 9.11 10 kg
e
e
e
hc m E h
E
m
− − − − −
− − −
× × ×
= − × = ×
×
× ×
= = ×
×
The energy of the electron is ~2.6eV, compared with 13.6eV for the electron held in the
1s orbital of H.
Monday assigned problems: Q13.3, P13.13, P13.17, P13.
Q13.3) Why can we conclude that the wave function
( ) ( , ) ( )
i E t
ψ x t ψ x e
−
= represents a
standing wave?
It represents a standing wave because it can be written as the product of a function
that depends only on time with a function that depends only on the spatial
coordinate. Therefore the nodes in ψ (^) ( x )do not move with time.
P13.13) Determine in each of the following cases if the function in the first column is an
eigenfunction of the operator in the second column. If so, what is the eigenvalue?
a) e
− i b 3 x + 2 y g ∂
2
2 x
b) x y
2 2
x y x x
c) sin θ cosθ
2 sin sin 6sin
d d
d d
a) (^) ( )
( 3 2 ) ,
i x y x y e
− + Ψ =
2
2
O ˆ
x
( )
( ) ( )
2 3 2 3 2 2
i x y e i x y O x y e x
− + ∂ − + Ψ = = − ∂
Eigenfunction with eigenvalue –9.
b)
2 2 x + y ( )
x y x x
( )
2 2 1 2 2 x y 2 2 x y x y x x
Eigenfunction with eigenvalue +1.
c) Ψ (^) ( θ ) = sin θ cosθ
sin sin 6 sin
d d O d d
( ) (^) ( )
( ) ( )
3 2 3
2 3
ˆ sin^ cos sin sin 6 sin cos sin sin 1 2 sin 6 sin cos
sin cos 6 sin cos 6 sin cos sin cos 1
d d d O d d d
Eigenfunction with eigenvalue +1.
P13.17) If two operators act on a wave function as indicated by AB f b g x , it is important
to carry out the operations in succession with the first operation being that nearest to the
function. Mathematically, AB f ^ (^) b g x (^) = A B f ^ d (^) b g x iand A ^ f x A A f ^ x
2 b g =^ d b gi. Evaluate the
following successive operations AB f (^) b g x. The operators A and B are listed in the first
two columns and f b g x is listed in the third column.
a)
d
d x
x x e
− a x^2
b) x
d
d x
x e
− a x^2
c) y x
x y
e
− a x e^2 + y^2 j
Note that your answers to parts (a) and (b) are not identical. As we will learn in Chapter
18, the fact that switching the order of the operators x and d dx changes the outcome of
the operation AB f (^) b g x is the basis for the Heisenberg uncertainty principle.
a) (^) ( )
2 2 3 2 2 2
d (^) a x a x a x x x e x e a x e d x
−^ = −^ −
