PHOTOELECTRIC EFFECT
Speed
of light (c) = 3x108 ms-1
Planck’s
constant (h) = 6.63 x 10-34 Js
Mass
of an electron = 9x10-31 kg
hf
= f + ½ mv2
wavelength = c / frequency
1. What is the
photoelectric effect?
Release of
photelectrons from the surface of metal upon illumination
2. What property must
the incoming radiation have to cause photoelectric emission from a surface?
Above threshold
value
3. What property of the
incoming radiation determines the energy of the emitted photoelectrons?
frequency
4. What property of the
incoming radiation determines the number of photoelectrons emitted per second?
intensity
5. Radiation of
wavelength 120 nm falls on a zinc plate. The threshold wavelength for this
surface is 200 nm.
(a) what is the energy of a quantum of the incident
radiation?
E = hf
(b) what is the energy required to give photoelectric
emission?
9.945e-19
(c) what is the energy of the emitted photoelectrons?
1.6575e-18 - 9.945e-19
6.1e-19
6. When radiation causes
photoelectric emission the electron emission is instantaneous - it occurs
immediately when radiation falls on the surface. What does this tell you about
the incident radiation?
Travels at the
speed of light
TAP 502-2: Photoelectric effect questions
hf = f + (1/2) mv2
and hf = f + eVs
e = 1.60 x 10-19 C,
h = 6.63 x 10-34 J s,
mass of electron = 9.11 x 10-31 kg
1 The work
function for lithium is 4.6 x 10-19 J.
(a) Calculate the
lowest frequency of light that will cause photoelectric emission.
(b) What is the
maximum energy of the electrons emitted when light of 7.3 x 1014 Hz
is used?
2 Complete the
table.
|
Metal
|
Work Function
/eV
|
Work Function
/J
|
Frequency
used /Hz
|
Maximum KE of
Ejected electrons /J
|
|
Sodium
|
2.28
|
|
6 x 1014
|
|
|
Potassium
|
|
3.68 x 10-19
|
|
0.32 x 10-19
|
|
Lithium
|
2.9
|
|
1 x 1015
|
|
|
Aluminium
|
4.1
|
|
|
0.35 x 10-19
|
|
Zinc
|
4.3
|
|
|
1.12 x 10-19
|
|
Copper
|
|
7.36 x 10-19
|
1 x 1015
|
|
3 The stopping
potential when a frequency of 1.61 x 1015 Hz is shone on a metal is
3 V.
(a) What is energy
transferred by each photon?
(b) Calculate the
work function of the metal.
(c) What is the
maximum speed of the ejected electrons?
4 Selenium has a work function of 5.11
eV. What frequency of light would just eject electrons? (The threshold
frequency is when the max KE of the ejected electrons is zero)
5 A frequency of
2.4 x 1015 Hz is used on magnesium with work function of 3.7 eV.
(a) What is energy
transferred by each photon?
(b) Calculate the
maximum KE of the ejected electrons.
(c) The maximum
speed of the electrons.
(d) The stopping
potential for the electrons.
Answers and worked solutions
1(a) hf
= f
hf = 4.60 x 10-19
f = 4.60 x 10-19 / 6.63 x 10-34
= 6.94 x 1014 Hz
(b) hf = f + (1/2) mv2. (6.63 x 10-34 x 7.30
x 1014) = 4.60 x 10-19 + (1/2) mv2
4.84 x 10-19
- 4.60 x 10-19 = (1/2) mv2
= 0.24 x 10-19 J
2
|
Metal
|
Work
Function
/
eV
|
Work
Function
/ J
|
Frequency
used
/ Hz
|
Maximum
KE of
ejected
electrons / J
|
|
Sodium
|
2.28
|
3.65 x 10-19
|
6 x
1014
|
0.35 x 10-19
|
|
Potassium
|
2.30
|
3.68
x 10-19
|
6 x 1014
|
0.32
x 10-19
|
|
Lithium
|
2.90
|
4.64 x 10-19
|
1.
x 1015
|
1.99 x 10-19
|
|
Aluminium
|
4.10
|
6.56 x 10-19
|
1.04 x 1015
|
0.35
x 10-19
|
|
Zinc
|
4.30
|
6.88 x 10-19
|
1.2 x 1015
|
1.12
x 10-19
|
|
Copper
|
4.60
|
7.36
x 10-19
|
1 x
1015
|
0
|
For copper 1 x 1015 Hz is below the threshold
frequency so no electrons are ejected.
3)
(a) 1.07
x 10-18 J
(b) hf
= f + eVs, so f = hf - eVs, so f = 1.07 x 10-18 –
(1.6 x 10-19 x 3) = 5.9 x 10-19
J
(c) eVs
= (1/2) mv2 so (1.60 x 10-19
x 3) = 0.5 x 9.11 x 10-31 x v2
so v2
=1.04 x 1012 and v = 1.02 x 106 m s-1
4 1.2 x 1015 Hz
5
(a) 1.6 x 10-18 J
(b) (1/2) mv2 = 1.0 x 10-18
J
(c) v2 = 1.1 x 1012 so
v = 1.1 x 106 m s-1
(d) (not
in spec) eVs = (1/2) mv2 so eVs = 1.00
x 10-18 and Vs = 0.63 V
