Съдържание
УсловиеPLANCK’S CONSTANT IN THE LIGHT OF AN INCANDESCENT LAMPDESCRIPTION OF THE APPARATUSSETTING UP THE EQUIPMENTTask 1 (2 т.)Task 2 — Measurement of the filament temperatureTask 3 — OPTICAL PROPERTIES OF THE FILTER (1 т.)Task 4 — PROPERTIES OF THE LDRAnswer sheet 1Answer sheet 2Answer sheet 3Answer sheet 4Answer sheet 5ОтговориРешениеIPhO 2005, Експеримент — Задача 1. Planck’s constant in the light of an incandescent lamp
36th International Physics Olympiad. Salamanca (España) 2005 — PLANCK’S CONSTANT IN THE LIGHT OF AN INCANDESCENT LAMP · 7 юли 2005 г. · 20 т.
Внимание: Бележка към темата
This is one connected experiment with four native tasks. Its 20-point total is the sum of marks printed on the five answer sheets. The complete original blank answer forms are preserved, including the intentionally blank headings on sheet 5. The separate multimeter and least-squares instruction page referenced in the apparatus list is absent from both archived question copies. The source’s numerical inconsistencies, including 0.51 versus 0.512, the voltage 100.1 in the sample table, differing resistance values and uncertainty substitutions, are retained as printed.
Условие
PLANCK’S CONSTANT IN THE LIGHT OF AN INCANDESCENT LAMP
In 1900 Planck introduced the hypothesis that light is emitted by matter in the form of quanta of energy . In 1905 Einstein extended this idea proposing that once emitted, the energy quantum remains intact as a quantum of light (that later received the name photon). Ordinary light is composed of an enormous number of photons on each wave front. They remain masked in the wave, just as individual atoms are in bulk matter, but – the Planck’s constant – reveals their presence. The purpose of this experiment is to measure Planck's constant.
A body not only emits, it can also absorb radiation arriving from outside. Black body is the name given to a body that can absorb all radiation incident upon it, for any wavelength. It is a full radiator. Referring to electromagnetic radiation, black bodies absorb everything, reflect nothing, and emit everything. Real bodies are not completely black; the ratio between the energy emitted by a body and the one that would be emitted by a black body at the same temperature, is called emissivity, , usually depending on the wavelength.
Planck found that the power density radiated by a body at absolute temperature in the form of electromagnetic radiation of wavelength can be written as

where and are constants. In this question we ask you to determine experimentally, which is proportional to .
For emission at small , far at left of the maxima in Figure F-1, it is permissible to drop the -1 from the denominator of Eq. (1), that reduces to
The basic elements of this experimental question are sketched in Fig. F-2.
- The emitter body is the tungsten filament of an incandescent lamp that emits a wide range of ’s, and whose luminosity can be varied.
- The test tube contains a liquid filter that only transmits a thin band of the visible spectrum around a value (see Fig. F-3). More information on the filter properties will be found in page 5.
- Finally, the transmitted radiation falls upon a photo resistor C (also known as LDR, the acronym of Light Dependent Resistor). Some properties of the LDR will be described in page 6.

The LDR resistance depends on its illumination, , which is proportional to the filament power energy density

where the dimensionless parameter is a property of the LDR that will be determined in the experiment. For this setup we finally obtain a relation between the LDR resistance and the filament temperature
that we will use in page 6. In this relation is an unknown proportionality constant. By measuring as a function on one can obtain , the objective of this experimental question.
DESCRIPTION OF THE APPARATUS
The components of the apparatus are shown in Fig. F-4, which also includes some indications for its setup. Check now that all the components are available, but refrain for making any manipulation on them until reading the instructions in the next page.

EQUIPMENT:
- Platform. It has a disk on the top that holds a support for the LDR, a support for the tube and a support for an electric lamp of 12 V, 0.1 A.
- Protecting cover.
- 10 turns and 1 kΩ potentiometer.
- 12 V battery.
- Red and black wires with plugs at both ends to connect platform to potentiometer.
- Red and black wires with plugs at one end and sockets for the battery at the other end.
- Multimeter to work as ohmmeter.
- Multimeter to work as voltmeter.
- Multimeter to work as ammeter.
- Test tube with liquid filter.
- Stand for the test tube.
- Grey filter.
- Ruler.
An abridged set of instructions for the use of multimeters, along with information on the least squares method, is provided in a separate page.
SETTING UP THE EQUIPMENT
Follow these instructions:
- Carefully make the electric connections as indicated in Fig. F-4, but do not plug the wires 6 to the potentiometer.
- By looking at Fig. F-5, follow the steps indicated below:

- Turn the potentiometer knob anticlockwise until reaching the end.
- Turn slowly the support for the test tube so that one of the lateral holes is in front of the lamp and the other in front of the LDR.
- Bring the LDR nearer to the test tube support until making a light touch with its lateral hole. It is advisable to orient the LDR surface as indicated in Fig. F-5.
- Insert the test tube into its support.
- Put the cover onto the platform to protect from the outside light. Be sure to keep the LDR in total darkness for at least 10 minutes before starting the measurements of its resistance. This is a cautionary step, as the resistance value at darkness is not reached instantaneously.
Task 1 (2 т.)
Task 1 Draw in Answer Sheet 1 the complete electric circuits in the boxes and between the boxes, when the circuit is fully connected. Please, take into account the indications contained in Fig. F-4 to make the drawings. [2 т.]
Task 2 — Measurement of the filament temperature
The electric resistance of a conducting filament can be given as
where is the resistivity of the conductor, is the length and the cross section of the filament.
This resistance depends on the temperature due to different causes such as:
- Metal resistivity increases with temperature. For tungsten and for temperatures in the range 300 K to 3655 K, it can be given by the empirical expression, valid in SI units,
- Thermal dilatation modifies the filament’s length and section. However, its effects on the filament resistance will be negligible small in this experiment.
From (4) and (5) and neglecting dilatations one gets
- Therefore, to get it is necessary to determine . This can be achieved by measuring the filament resistance at ambient temperature .
a) Measure with the multimeter the ambient temperature . [1 т.]
b) It is not a good idea to use the ohmmeter to measure the filament resistance at because it introduces a small unknown current that increases the filament temperature. Instead, to find connect the battery to the potentiometer and make a sufficient number of current readings for voltages from the lowest values attainable up to 1 V. (It will prove useful to make at least 15 readings below 100 mV.) At the end, leave the potentiometer in the initial position and disconnect one of the cables from battery to potentiometer.
Find for each pair of values of and , translate these values into the Table for Task 2,b) in the Answer Sheets. Indicate there the lowest voltage that you can experimentally attain. Draw a graph and represent in the vertical axis against . [2 т.]
c) After inspecting the graphics obtained at b), select an appropriate range of values to make a linear fit to the data suitable for extrapolating to the ordinate at the origin, . Write the selected values in the Table for Task 2, c) in the Answer Sheets. Finally, obtain and . [2,5 т.]
d) Compute the numerical values of and for in Ω and in K using (6). [1 т.]
Task 3 — OPTICAL PROPERTIES OF THE FILTER (1 т.)
The liquid filter in the test tube is an aqueous solution of copper sulphate (II) and Orange (II) aniline dye. The purpose of the salt is to absorb the infrared radiation emitted by the filament.
The filter transmittance (transmitted intensity/incident intensity) is shown in Figure F-6 versus the wavelength.

Task 3 Determine and from Fig. F-6. [1 т.]
Note: is the total width at half height and the wavelength at the maximum.
Task 4 — PROPERTIES OF THE LDR
The material which composes the LDR is non conducting in darkness conditions. By illuminating it some charge carriers are activated allowing some flow of electric current through it. In terms of the resistance of the LDR one can write the following relation
where is a constant that depends on the composition and geometry of the LDR and is a dimensionless parameter that measures the variation of the resistance with the illumination produced by the incident radiation. Theoretically, an ideal LDR would have , however many factors intervene, so that in the real case .

It is necessary to determine . This is achieved by measuring a pair and (Fig. F-7) and then introducing between the lamp and the tube the grey filter F (Fig. F-8) whose transmittance is known to be 51.2 %, and we consider free of error. This produces an illumination . After measuring the resistance corresponding to this illumination, we have

From this
Do not carry out this procedure until arriving at part b) of task 4 below.
a) Check that the LDR remained in complete darkness for at least 10 minutes before starting this part. Connect the battery to the potentiometer and, rotating the knob very slowly, increase the lamp voltage. Read the pairs of values of and for in the range between 9.50 V and 11.50 V, and obtain the corresponding LDR resistances . (It will be useful to make at least 12 readings). Translate all these values to a table in the Answer Sheet. To deal with the delay in the LDR response, we recommend the following procedure: Once arrived at V, wait 10 min approximately before making the first reading. Then wait 5 min for the second one, and so on. Before doing any further calculation go to next step. [2 т.]
b) Once obtained the lowest value of the resistance , open the protecting cover, put the grey filter as indicated in F-9, cover again - as soon as possible - the platform and record the new LDR resistance . Using these data in (8) compute and .

[1,5 т.]
c) Modify Eq. (3) to display a linear dependence of ln on . Write down that equation there and label it as (9). [1 т.]
d) Using now the data from a), work out a table that will serve to plot Eq. (9). [3 т.]
e) Make the graphics plot and, knowing that , compute and by any method (you are allowed to use statistical functions of the calculators provided by the organization). [3 т.]
(Speed of light, ; Boltzmann constant, )
Answer sheet 1
Draw the electric connections in the boxes and between boxes below.

Answer sheet 2
TASK 2
a) (1.0 points)
b) (2.0 points)
, ,
*
- This is a characteristic of your apparatus. You can’t go below it.

Answer sheet 3
TASK 2
c) (2.5 points)
, ,
,
d) (1.0 points)
,
TASK 3 (1.0 points)
,

Answer sheet 4
TASK 4
a) (2.0 points)
, ,
b) (1.5 points)
, , ,
c) (1.0 points)
Eq. (9)

Answer sheet 5
TASK 4
d) (3.0 points)
, ,
e) (3.0 points)
,

Отговори
Task 1, Task 1 The official answer is the connection diagram (solutions, page 1).
Task 2 — Measurement of the filament temperature, a) 297 K
Sample value from the official solution ( ºC); it depends on the laboratory.
Task 2 — Measurement of the filament temperature, c) 11.4 Ω
Sample value from the official solution: , .
Task 2 — Measurement of the filament temperature, d) 39.4
Official sample value , (SI, in Ω and in K).
Task 3 — OPTICAL PROPERTIES OF THE FILTER, Task 3 λ0 = 590 nm, Δλ = 28 nm
Official solution: , nm, nm.
Task 4 — PROPERTIES OF THE LDR, b) 0.702
Official sample values: kΩ, kΩ, , .
Task 4 — PROPERTIES OF THE LDR, c)
Task 4 — PROPERTIES OF THE LDR, e) J·s
Official sample result: J · s, J · s.
Решение
Task 1
TASK 1
Draw the electric connections in the boxes and between boxes below.

Task 2
TASK 2
a)
b)
| /mV | / mA | /Ω |
|---|---|---|
| 21.9 | 1.87 | 11.7 |
| 30.5 | 2.58 | 11.8 |
| 34.9 | 2.95 | 11.8 |
| 37.0 | 3.12 | 11.9 |
| 40.1 | 3.37 | 11.9 |
| 43.0 | 3.60 | 11.9 |
| 47.6 | 3.97 | 12.0 |
| 51.1 | 4.24 | 12.1 |
| 55.3 | 4.56 | 12.1 |
| 58.3 | 4.79 | 12.2 |
| 61.3 | 5.02 | 12.2 |
| 65.5 | 5.33 | 12.3 |
| 67.5 | 5.47 | 12.3 |
| 73.0 | 5.88 | 12.4 |
| 80.9 | 6.42 | 12.6 |
| 85.6 | 6.73 | 12.7 |
| 89.0 | 6.96 | 12.8 |
| 95.1 | 7.36 | 12.9 |
| 111.9 | 8.38 | 13.4 |
| 130.2 | 9.37 | 13.9 |
| 181.8 | 11.67 | 15.6 |
| 220 | 13.04 | 16.9 |
| 307 | 15.29 | 20.1 |
| 447 | 17.68 | 25.1 |
| 590 | 19.8 | 29.8 |
| 730 | 21.5 | 33.9 |
| 860 | 23.2 | 37.1 |
| 960 | 24.4 | 39.3 |
= 9.2 mV *
* This is a characteristic of your apparatus. You can´t go below it.
We represent in the vertical axis against .

In order to work out , we choose the first ten readings.
TASK 2
c)
| /mV | / mA | /Ω |
|---|---|---|
| 21.9 ± 0.1 | 1.87 ± 0.01 | 11.7 ± 0.1 |
| 30.5 ± 0.1 | 2.58 ± 0.01 | 11.8 ± 0.1 |
| 34.9 ± 0.1 | 2.95 ± 0.01 | 11.8 ± 0.1 |
| 37.0 ± 0.1 | 3.12 ± 0.01 | 11.9 ± 0.1 |
| 40.1 ± 0.1 | 3.37 ± 0.01 | 11.9 ± 0.1 |
| 43.0 ± 0.1 | 3.60 ± 0.01 | 11.9 ± 0.1 |
| 47.6 ± 0.1 | 3.97 ± 0.01 | 12.0 ± 0.1 |
| 51.1 ± 0.1 | 4.24 ± 0.01 | 12.1 ± 0.1 |
| 55.3 ± 0.1 | 4.56 ± 0.01 | 12.1 ± 0.1 |
| 58.3 ± 0.1 | 4.79 ± 0.01 | 12.2 ± 0.1 |

Error for (We work out the error for first value, as example).
We have worked out by the least squares.
For axis :
For axis :
d)
Working out the error for two methods:
Method A
Method B
Higher value of :
Smaller value of :
Task 3
TASK 3
Because of
| nm | nm |
|---|
Task 4
TASK 4
a)
| /V | / mA | /kΩ |
|---|---|---|
| 9.48 | 85.5 | 8.77 |
| 9.73 | 86.8 | 8.11 |
| 9.83 | 87.3 | 7.90 |
| 100.1 | 88.2 | 7.49 |
| 10.25 | 89.4 | 7.00 |
| 10.41 | 90.2 | 6.67 |
| 10.61 | 91.2 | 6.35 |
| 10.72 | 91.8 | 6.16 |
| 10.82 | 92.2 | 6.01 |
| 10.97 | 93.0 | 5.77 |
| 11.03 | 93.3 | 5.69 |
| 11.27 | 94.5 | 5.35 |
| 11.42 | 95.1 | 5.17 |
| 11.50 | 95.5 | 5.07 |
b)
Because of
For working out we know that:
Transmittance,
Working out the error for two methods:
Method A
Method B
Higher value of :
Smaller value of :
| kΩ | |
|---|---|
| kΩ |
c)
We know that (3)
then
Because of (6)
consequently
d)
| /V | / mA | / Ω | / K | (S.I.) | / kΩ | ln |
|---|---|---|---|---|---|---|
| 9.48 ± 0.01 | 85.5 ± 0.1 | 110.9 ± 0.2 | 1962 ± 18 | 8.77 ± 0.01 | 2.171 ± 0.001 | |
| 9.73 ± 0.01 | 86.8 ± 0.1 | 112.1 ± 0.2 | 1980 ± 18 | 8.11 ± 0.01 | 2.093 ± 0.001 | |
| 9.83 ± 0.01 | 87.3 ± 0.1 | 112.6 ± 0.2 | 1987 ± 18 | 7.90 ± 0.01 | 2.067 ± 0.001 | |
| 10.01 ± 0.01 | 88.2 ± 0.1 | 113.5 ± 0.2 | 2000 ± 18 | 7.49 ± 0.01 | 2.014 ± 0.001 | |
| 10.25 ± 0.01 | 89.4 ± 0.1 | 114.7 ± 0.2 | 2018 ± 18 | 7.00 ± 0.01 | 1.946 ± 0.001 | |
| 10.41 ± 0.01 | 90.2 ± 0.1 | 115.4 ± 0.2 | 2028 ± 18 | 6.67 ± 0.01 | 1.894 ± 0.002 | |
| 10.61 ± 0.01 | 91.2 ± 0.1 | 116.3 ± 0.2 | 2041 ± 18 | 6.35 ± 0.01 | 1.849 ± 0.002 | |
| 10.72 ± 0.01 | 91.8 ± 0.1 | 116.8 ± 0.2 | 2049 ± 19 | 6.16 ± 0.01 | 1.818 ± 0.002 | |
| 10.82 ± 0.01 | 92.2 ± 0.1 | 117.4 ± 0.2 | 2057 ± 19 | 6.01 ± 0.01 | 1.793 ± 0.002 | |
| 10.97 ± 0.01 | 93.0 ± 0.1 | 118.0 ± 0.2 | 2066 ± 19 | 5.77 ± 0.01 | 1.753 ± 0.002 | |
| 11.03 ± 0.01 | 93.3 ± 0.1 | 118.2 ± 0.2 | 2069 ± 19 | 5.69 ± 0.01 | 1.739 ± 0.002 | |
| 11.27 ± 0.01 | 94.5 ± 0.1 | 119.3 ± 0.2 | 2085 ± 19 | 5.35 ± 0.01 | 1.677 ± 0.002 | |
| 11.42 ± 0.01 | 95.1 ± 0.1 | 120.1 ± 0.2 | 2096 ± 19 | 5.15 ± 0.01 | 1.639 ± 0.002 | |
| 11.50 ± 0.01 | 95.5 ± 0.1 | 120.4 ± 0.2 | 2101 ± 19 | 5.07 ± 0.01 | 1.623 ± 0.002 | |
| unnecessary |
We work out the errors for all the first row, as example.
Error for :
Error for :
Error for :
Error for ln :
e)
We plot ln versus .

By the least squares
For axis :
For axis :
Because of
and
then
| J · s | J · s |
|---|
Оригинал в Архива: exp.pdf · официални решения: exp_sol.pdf · допълнителни данни: IPhO_2005_Exp_question(pdf).pdf · допълнителни данни: IPhO_2005_Exp_solution(pdf).pdf