Le Fevre High School
SACE Stage 2 Physics
Photons
1
(a) f = 6.5 x 1014 Hz since E = hf = (6.63 x 10-34)(6.5 x 1014) = 4.3 x 10-19 J (or 2.69 eV)
Since p = mc =
4.3 x 10 19
mc 2
E
hf
h
=
=
or  p =
= 1.43 x 10-27 kg m s-1
c
c
c

3 x 10 8
(b)  = 4.5 x 10-11 m  f 
3 x 10 8
c
=
= 6.67 x 1018 Hz

4.5 x10 11
E = hf = (6.63 x 10-34)(6.67 x 1018) = 4.42 x 10-15 J (or 2.76 x 10-4 eV)
(6.63 x 10 34 )
mc 2
E
hf
h
=
=
or  p =
= 1.47 x 10-23 kg m s-1
c
c
c

4.5 x 10 11
E
(c) p = 2.0 x 10-27 kg m s-1 and as p =  E = pc = (2.0 x 10-27)( 3.0 x 108) = 6 x 10-19 J (or
c
Since p = mc =
3.75 eV)
2
E = nhf = (1020)(6.63 x 10-34)(5.5 x 1014) = 36.47 J (or 2.28 x 1020 eV)
3
E
 E = Pt  E = (2 x 10-3)(1) = 2 x 10-3 J
t
(2 x 10 3 )(5.5 x 10 7 )
E
c
hence n =
and as  =  n =
= 5.4 x 1015 photons
f
hf
(6.63 x 10 34 )(3 x 10 8 )
P=
4.
(a) Einstein proposed a quantised model light. Light consists of quantised bundles of energy with
wave properties. The energy of the quanta or photon of light is directly proportional to the
frequency of the light. A metal has large numbers of "free" surface electrons, bound to the metal
surface but not a particular atom. The most energetic "free" electrons requires the minimum
energy (or Work Function) to escape the metal' attraction. If one photon interacts with an
electron the photon may be absorbed and if the energy is sufficient the electron overcomes the
binding energy of the metal and escapes with excess kinetic energy.
Thus by conservation of energy
kinetic energy = Energy of photon - Binding energy , or for the most energetic "free" electrons
Kmax = hf - W
Hence if the photon has sufficient energy (greater than the threshold frequency) then electrons
escape from the surface of a metal when light is incident on the surface.
Le Fevre High School
(a) Work function = hfo = 6.63 x 10-34 x 5 x 1014 = 3.3 x 10-19 J
Now an energy of 1 electron-volt is that energy gained by an electron passing through a
potential difference of 1 volt,
i.e. 1 eV = 1.6 x 10-19 joule
Hence, work function (in eV) =
3.31x10 19
= 2.07 eV
1.6x10 19
= 2.1 eV
(b) From Einstein's Photoelectric equation, we have
1
mv2max = hf - hfo = h(f - fo)
2
= 6.63 x 10-34 (8 x 1014-5 x 1014)
= 6.62 x 10-20 x 3 joule

v2max =

5. (a)
2x6.62x10 20 x3
9x10 31
vmax = 6.6 x 105 m s-1
Slope is constant (depends on h only)
X intercept depends on work function of metal (W = hfo), therefore independent of intensity.
Hence the graph required has the same slope and same threshold frequency, fo i.e. graph B
(b) Work function twice as large, therefore fo twice as large.
Constant slope, therefore C.
6
fo the threshold
frequency
Max. Kinetic Energy (eV)
Photo-electric effect
3
2
1
0
-1 0
2
4
6
8
10
12
14
-2
-3
Frequency (x 1014 Hz)
Choose two points on the graph (MUST lie on the line of best fit) e.g. (5.6 x 1014, 0.12) and (11.2 x
1014, 2.46) It is also best to select points at the ends of the line to reduce the percentage error.
(a) Read threshold frequency from the graph
fo = 5.3 x 1014 Hz
(b) Work function is the Y intercept = 2.3 eV (approximately)
NB cannot use hfo = (6.63 x 10-34)(53. x 1014) as 'h' is not from the graph. You
could do part (c), find 'h' and then do part (b).
Le Fevre High School
y 2 y1
x 2 x 1
2.440.12
=
(11.25.6)x1014
(c) Plank's constant = slope of graph =
= 4.14 x 1015 eV s.
To convert to joule second multiply by 1.6 x 10-19
i.e. 4.14 x 1015 x 1.6 x 1019 = 6.67 x 10-34 J s
i.e. Answer 6.7 x 10-34 J s
(OR convert eV to Joule before you start.)
7
(a)
E = hf = 6.63 x 10-34 x 7.0 x 1014 = 4.64 x 10-19 J
(b)
W
(c)
Kmax = hf - W = hf - hfo = (4.64 - 1.99) x 10-19 = 2.65 x 10-19 J
(d)
1
2E
mv2 = E  v2 =
and
2
m
v=
(e)
8.
W=
2 x 2.65 x 10 19
2E
=
9.11 x 10  31
m
= 7.6 x 105 m s-1
199
. x 10 19
= 1.24 eV
16
. x 10 19
hc
W = hfo =
 max
 max =
hc
W
=
= hfo = 6.63 x 10-34 x 3.0 x 1014 = 1.99 x 10-19 J
6.63 x 10 34 x 3 x 10 8
4.5 x 16
. x 10 19
= 2.8 x 10-7 m
9. Number of radio photons =
E
E radio
=
hf 
hf r
10. 5 eV = 5 x 1.6 x 10-19J = 8 x 10-19 J
10 20
=
=
10 6
1014
Le Fevre High School
3 x 10 8
f=
=
= 1.18 x 1015 Hz

2.536 x 10  7
c
11. (a) KEmax = 0.12 eV
(b) Kmax = hf - W
W
= hf - Kmax
= 4.13 x 10-15 x
3x10 8
- 0.12eV
2.536 x10 7
= 4.9 -0.12
= 4.78 eV
12. (a) E
= hf =
hc

= (6.63.x10-34 x 3 x 108)/(3.75 x 10-7)
= 5.3 x 10-19 J
To convert to electron volts, divide by 1.6 x 10-19
E = 3.3 eV
(b) Kmax = hf - W
= 3.3-2.0 = 1.3 eV
Kmax= 2.08 x 10-19 J
2 x 2.08 x 10 19
25
=
9.11 x 10  31
m
= 6.8 x 105 m s-1
(c) v =