Worked examples for calculations regarding the photoelectric effect

1. Photoelectric Effect
Calculations

2. ENERGY PER PHOTON
E=hf

3. E = hf
• E = Energy per photon
• f = frequency of light
• h = Plank’s constant = 6,63 x 10-34 J·s

4. Question 1
How much energy does each photon of UV light
of wavelength 288 nm have?

5. Solution: Given and required
Required: E
Given: 𝜆 = 288 nm
Known: h = 6,63 x 10-34 J·s
Equation: E = hf

6. 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 =
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡
𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡
𝑓 =
𝑐
𝜆
𝑓 =
3 × 108
𝑚 ∙ 𝑠−1
𝜆
Solution: how to find f

7. Solution: finding f
E = hf
𝑓 =
3 × 108 𝑚 ∙ 𝑠−1
𝜆
Given: 𝜆 = 288 nm
Known: nm = 10-9 m
Therefore 288 nm = 288 x 10-9 m

8. Solution: finding f
E = hf
𝑓 =
3 × 108
𝑚 ∙ 𝑠−1
𝜆
𝑓 =
3 × 108
𝑚 ∙ 𝑠−1
288 × 10−9 𝑚
Given: 𝜆 = 288 nm
Known: nm = 10-9 m
Therefore 288 nm = 288 x 10-9 m
s-1 =
1
𝑠
= Hz

9. Solution: finding f
E = hf
𝑓 =
3 × 108
𝑚 ∙ 𝑠−1
𝜆
𝑓 =
3 × 108
𝑚 ∙ 𝑠−1
288 × 10−9 𝑚
f = 1,04 x 1015 Hz

10. Solution: finding E
E = hf
Required: E
Known: h = 6,63 x 10-34 J·s
f = 1,04 x 1015 Hz

11. Solution: finding E
E = hf
= 6,63 x 10-34 J·s x 1,04 x 1015 Hz
= 6,63 x 10-34 J·s x 1,04 x 1015 s-1
= 6,91 x 10-19 J

12. PHOTON

13. PHOTON
carries amount of
Energy = hf

14. PHOTON
carries amount of
Energy ∝ f

15. PHOTON
carries amount of
Energy = hf

16. PHOTON
carries amount of
Energy = hf
- - - - - - - -
givesALLtoONE
in
ELECTRON metal

17. in
metal
PHOTON
carries amount of
Energy = hf
- - - - - - - -
givesALLtoONE
ELECTRON

18. PHOTON
carries amount of
Energy = hf
- - - - - - - -
givesALLtoONE
in
ELECTRON metal
photoelectric emission
-

19. - - - - - - - -
PHOTON
carries amount of
givesALLtoONE
in
Energy = hf
photoelectric emission
-
ELECTRON
type determinesneeds ≥
for
W0 = hf0
metal

20. type determines
metal
- - - - - - - -
PHOTON
carries amount of
givesALLtoONE
in
Energy = hf
-
needs ≥
for
W0 = hf0
ELECTRON
photoelectric emission

21. - - - - - - - -
PHOTON
carries amount of
givesALLtoONE
in
Energy = hf
photoelectric emission
-
ELECTRON
type determinesneeds ≥
for
W0 = hf0
metal

22. W0 = hf0
type determines
metalin
carries amount of
Energy = hf
needs ≥- - - - - - - -
givesALLtoONE
-
ELECTRON
for
photoelectric emission
PHOTON

23. in
type determinesneeds ≥
for
metal
- - - - - - - -
PHOTON
carries amount of
givesALLtoONE
Energy = hf
photoelectric emission
-
ELECTRON
W0 = hf0

24. W0 = hf0
in
type determinesneeds ≥
for
metal
- - - - - - - -
carries amount of
givesALLtoONE
photoelectric emission
-
ELECTRON
Energy = hf
PHOTON

25. type determines
- - - - - - - -
PHOTON
carries amount of
givesALLtoONE
in
Energy = hf
photoelectric emission
-
needs ≥
for
W0 = hf0
ELECTRON metal

26. type determines
- - - - - - - -
PHOTON
carries amount of
givesALLtoONE
in
Energy = hf
photoelectric emission
-
needs ≥
for
W0 = hf0
ELECTRON metal
for surface electrons occurs with
KEmax= hf–hf0

27. type determines
- - - - - - - -
PHOTON
carries amount of
givesALLtoONE
in
Energy = hf
photoelectric emission
-
needs ≥
for
W0 = hf0
ELECTRON metal
for surface electrons occurs with
KEmax= hf–hf0
KEmax= ½ mvmax
2
=

28. type determines
- - - - - - - -
PHOTON
carries amount of
givesALLtoONE
in
Energy = hf
photoelectric emission
-
needs ≥
for
W0 = hf0
ELECTRON metal
for surface electrons occurs with
KEmax= hf–hf0
KEmax= ½ mvmax
2
=

29. energy photon gives
energy e- needs to
break free from Zn

30. energy photon gives
=6,91 x 10-19 J
energy e- needs to
break free from Zn

31. energy photon gives
=6,91 x 10-19 J
energy e- needs to
break free from Zn
=6,91 x 10-19 J

32. energy photon gives
=6,91 x 10-19 J
Zn’s work function, W0
=6,91 x 10-19 J

33. energy photon gives
energy e-
needs to
break free
e- ‘s
kinetic
energy

34. E =hf
W0=hf0 KE

35. E =hf
W0=hf0 KE

36. E =hf
W0=hf0 KE

37. Metal 𝝀0 (nm) f0 (Hz) Movie
Pt 196 1,53 x 1015
Cu 263 1,14 x 1015
Zn 289 1,04 x 1015
Ca 427 7,03 x 1014
Na 539 5,57 x 1014

38. Na’s W0

39. Na’s W0
=3,38 x 10-19 J

40. Zn’s W0
=6,91 x 10-19 J
Na’s W0
=3,38 x 10-19 J

41. Question 2
How much energy does each photon of UV light
of frequency 1,5 x 1015 Hz have?

42. Solution
E = hf
= 6,63 x 10-34 J·s x 1,5 x 1015 Hz
= 6,63 x 10-34 J·s x 1,5 x 1015 s-1
= 9,95 x 10-19 J

43. energy e- needs to
break free from Zn
=6,91 x 10-19 J

44. energy photon gives
=9,95 x 10-19 J
energy e- needs to
break free from Zn
=6,91 x 10-19 J

45. energy photon gives
energy e- needs to
break free from metal
e- ‘s
kinetic
energy

46. E=hf
energy e- needs to
break free from metal
e- ‘s
kinetic
energy

47. E=hf
W0=hf0 e- ‘s
kinetic
energy

48. E=hf
W0=hf0 KE = E-W0
hf-hf0

49. Question 3
How much kinetic energy do electrons emitted
from the surface of zinc have when light of
frequency 1,5 x 1015 Hz is incident on the zinc?

50. Solution
KE = E-W0
= 9,95 x 10-19 J - 6,91 x 10-19 J
= 3,03 x 10-19 J

51. E=hf
=9,95 x 10-19 J

52. E=hf
=9,95 x 10-19 J
W0=hf0
=6,91 x 10-19 J
e- ‘s
kinetic
energy

53. E=hf
=9,95 x 10-19 J
W0=hf0
=6,91 x 10-19 J
KE = hf-W0
= 3,03 x 10-19 J

54. Question 4
At what maximum speed are electrons emitted
from the surface of zinc have when light of
frequency 1,5 x 1015 Hz is incident on the zinc?

55. Solution: v subject of formula
KEmax = ½ mvmax
2
KEmax x
2
𝑚
= ½ mvmax
2 x
2
𝑚
2KEmax
𝑚
= vmax
2
2KEmax
𝑚
= vmax
2
2KEmax
𝑚
= 𝑣
𝑣 𝑚𝑎𝑥=
2KEmax
𝑚

56. Solution: Known values
𝑣 𝑚𝑎𝑥=
2KEmax
𝑚
Known: m of 1 e- = 9,11 x 10-31 kg
Known: 3,03 x 10-19 J (from Question 3)

57. Solution
𝑣 𝑚𝑎𝑥 =
2(3,03 x 10−19)
9,11 x 10−31
𝑣 𝑚𝑎𝑥= 815 599 m•s−1
𝑣 𝑚𝑎𝑥= 816 000 m•s-1

58. Question 4
Light of a certain frequency is incident on a
piece of sodium. The maximum kinetic energy of
the emitted electrons is 1,07x10-20 J. What is the
frequency of the incident light?

59. Solution: making f the subject of the
formula
KEmax = E - W0
KEmax = hf – hf0
KEmax + hf0= hf – hf0 + hf0
KEmax + hf0= hf
hf = KEmax + hf0
ℎ𝑓
ℎ
=
KEmax + hf0
ℎ
𝑓 =
KEmax + hf0
ℎ

60. Solution: Known values
𝑓 =
KEmax + hf0
ℎ
KEmax = 1,07x10-20 J
Na’s f0 = 5,57 x 1014 Hz
h = 6,63 x 10-34 J·s

61. Solution: Known values
𝑓 =
1,07x10−20 + 6,63 x 10−34•5,57 x 1014
6,63 x 10−34
f = 5,73 x 1014 Hz

62. energy photon gives
energy e-
needs to
break free
e- ‘s
kinetic
energy

63. E =hf
W0=hf0 KE

64. E =hf
W0=hf0 KE

65. E =hf
W0=hf0 KE

66. E =hf
W0=hf0 KE

67. LIGHT
colour
has
photoelectric emission
determined
by light’s
frequency
PHOTONSconsists
of
energy
carry amount of
determined by
determines if
occurs from a
=f0<f0 >f0
if if
type
determines f0W0
W0
for emission,
each e- needs energy ≥
has has >
is an
amount of
metal

68. LIGHT
intensity
photoelectric emission
determined by
emission rate
PHOTONSconsists of
metal
occurs from a
of
affects rate that
has

69. LIGHT
intensity
colour
has
photoelectric emission
determined
by light’s
frequency
determined by
emission rate
PHOTONSconsists of
energy
carry amount of
determined by
metal
determines if
occurs from a
of
affects rate that

70. LIGHT
intensity colour
has
photoelectric emission
determined by light’s
frequency
determined by
emission rate
PHOTONSconsists of
energy
carry amount of
determined by
metal
determines if
occurs
from a
of
affects rate of