Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2010
Laura Gagliardi
Answers to Homework Set 1
To be collected on September 17 2010
From lecture 1: What is the maximum number of electrons that can be emitted if a
potassium surface of work function 2.45 eV absorbs 3.20 x 10
–3
J of radiation at a
wavelength of 250 nm? What is the kinetic energy and velocity of the emitted electrons?
Frequency
ν
is related to wavelength
λ
by the equation
c
=
λν
where
c
is the speed of light (2.998
x
10
8
m sec
–1
). Thus, the frequency of the incident
light is
The energy of photons having this frequency is computed from the equation
E
=
h
ν
where
h
is Planck’s constant (6.626
x
10
–34
J s). Carrying out the relevant multiplication
gives
E
= 7.946
x
10
–19
J. Using the conversion factor 1 J = 6.24
x
10
18
eV gives an
energy of 4.95 eV. If the work function is 2.45 eV, this leaves 2.50 eV (4.021
x
10
–19
J) of
energy that is converted into electron kinetic energy. To determine the velocity of the
electrons, we use
v
=
2
T
m
e
=
2
×
4.021
×
10
−
19
J
9.109
×
10
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 Fall '08
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 Physical chemistry, Electron, pH, Energy, Kinetic Energy, Light, ev

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