Ex.9.1
3. A light meter registers a value of 100 W m-2 when placed at a distance of 50 cm from a small bright light in a dark room.
a) What value will it register at a distance of 150 cm from the same source? (2)
I1 d12 = I2 d22
100 x 502 = I2 x 1502
I2 = 11.1 W m-2
b) What value will it register at a distance of 25 cm from the same source? (2)
I1 d12 = I2 d22
100 x 502 = I2 x 252
I2 = 400 W m-2
Ex.9.2
6. Calculate the energy of a photon of light of wavelength 405 nm. (2)
v = f λ
3 x 108 = f x 405 x 10-9
f = 7.14 x 1014 Hz
E = h f
E = 6.63 x 10-34 x 7.41 x 1014
E = 4.9 x 10-19 J
7. Calculate the wavelength of a photon of energy 4.2 x 10-19 J. (2)
E = h f
4.2 x 10-19 = 6.63 x 10-34 x f
f = 6.33 x 1014 Hz
v = f λ
3 x 108 = 6.33 x 1014 x λ
λ = 4.74 x 10-7 m
10. The work function for sodium metal is 2.9 x 10-19 J.
Light of wavelength 5.4 x 10-7 m strikes the surface of this metal.
What is the maximum kinetic energy of the electrons emitted from the surface? (3)
v = f λ
3 x 108 = f x 5.4 x 10-7
f = 5.56 x 1014 Hz
For the photon
E = h f
E = 6.63 x 10-34 x 5.56 x 1014
E = 3.67 x 10-19 J
Now use
E = φ + Ek
3.67 x 10-19 = 2.9 x 10-19 + Ek
Ek = 7.8 x 10-20 J
Ex.9.3
2.
The diagram shows the energy levels for the hydrogen atom.
a) Between which two energy levels would an electron transition lead to emission of radiation of highest frequency? (1)
From E4 to E0
b) Calculate the frequency of this radiation. (3)
ΔE = (21.76 - 0.864) x 10-19 = 2.0896 x 10-18 J
E = h f
2.0896 x 10-18 = 6.63 x 10-34 x f
f = 3.15 x 1015 Hz
Ex.9.4
1. The emission spectrum of helium contains only a few sharp lines of light in certain colours.
Explain how these lines are produced. (2)
As an electron makes the transition to a lower energy level a photon of radiation is released corresponding to the energy given up by the electron.
Only certain transitions are allowed.
11.
A Bunsen flame containing vaporised sodium is placed between a sodium vapour lamp and a screen.
a) Explain why the flame casts a dark shadow. (2)
Photons from the sodium lamp are readily absorbed by the vaporised sodium. These photons have the exact amount of energy required for the electrons in a sodium atom to move to an excited state.
b) The sodium lamp is now replaced with a cadmium lamp.
Explain why there is no longer a dark shadow on the screen. (2)
The photons from the cadmium atoms do not have the exact amount of energy corresponding to an energy transition in sodium. They are not absorbed so pass on to illuminate the screen.
Past Papers
3.
A student directs a ray of monochromatic light through a semicircular glass block.
At point X the incident ray splits into two rays:
T - a transmitted ray and
R - a reflected ray.
The student uses a light meter to measure the irradiance of ray R as angle θ is changed.
(a) State what is meant by the irradiance of a radiation. (1)
Irradiance is power per unit area: I = P / A.
(b) Explain why, as angle θ is changed, it is important to keep the light meter at a constant distance from point X for each measurement of intensity. (1)
Changing the distance d may change the irradiance. e.g. For a point source I α 1 / d2.
By keeping the distance constant it will not affect irradiance.
(c) The graph below is obtained from the student's results.
graph.
(i) What is the value of the critical angle in the glass for this light? (1)
42°
(ii) Calculate the refractive index of the glass for this light. (2)
n = 1 / sinθc
n = 1 / sin42° = 1.49
(iii) As the angle θ is increased, what happens to the intensity of ray T? (1)
The irradiance of the transmitted ray T decreases.
Note that the graph is for ray R not T!
4.
The apparatus shown is used to investigate photoelectric emission from a metal surface.
The irradiance and frequency of the incident radiation can be varied as required.
(a) (i) Explain what is meant by photoelectric emission from a metal. (2)
Photons eject electrons from the surface of a metal. They must have more energy than the work function.
(ii) What name is given to the minimum frequency of the radiation that produces a current in the circuit? (1)
Threshold frequency.
(iii) A particular source of radiation produces a current in the circuit.
Explain why the current increases as the irradiance of the radiation increases. (1)
More photons with enough energy are incident on the metal releasing electrons.
From I = N h f. Where N is the number of photons per second per square meter. As N increases so does the number of ejected electrons.
(b) A semiconductor chip is used to store information. The information can only be erased by exposing the chip to ultraviolet radiation for a period of time.
The following data is provided:
Frequency of ultraviolet radiation used = 9.0 x 1014 Hz
Minimum irradiance of UV required at chip = 25 W m-2
Area of chip exposed to radiation = 1.8 x 10-9 m2
Time taken to erase the information = 15 minutes
Energy of radiation needed to erase the information = 40.5 µJ
(i) Calculate the energy of a photon of the ultraviolet radiation used. (2)
E = h f = 6.63 x 10-34 x 9 x 1014 = 5.97 x 10-19 J
(ii) Calculate the number of photons of the ultraviolet radiation required to erase the information. (2)
number of photons = total energy / energy of one photon
40.5 x 10-6 / 5.97 x 10-19 = 6.79 x 1013 photons
(iii) Sunlight of irradiance 25 W m-2, at the chip, can also be used to erase the information.
State whether the time taken to erase the information is greater than, equal to or less than 15 minutes.
You must justify your answer. (1)
It will take a longer time. Only some of the photons in sunlight are ultraviolet. The rest are visible or infrared photons and individually will not have enough energy to contribute to erasing the chip.
Total marks = (36)