IPhO 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 hνh\nu. 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 hh – 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, ε\varepsilon, usually depending on the wavelength.

Planck found that the power density radiated by a body at absolute temperature TT in the form of electromagnetic radiation of wavelength λ\lambda can be written as

uλ=εc1λ5(ec2/λT−1)(1)u_\lambda = \varepsilon\frac{c_1}{\lambda^5\left(e^{c_2/\lambda T}-1\right)} \qquad (1)
Sketch of the power density u_λ versus wavelength λ for three temperatures T1 < T2 < T3; each curve rises steeply to a maximum and then decays, the maximum being higher and further left for higher temperature.

where c1c_1 and c2c_2 are constants. In this question we ask you to determine c2c_2 experimentally, which is proportional to hh.

For emission at small λ\lambda, 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

uλ=εc1λ5ec2/λT(2)u_\lambda = \varepsilon\frac{c_1}{\lambda^5 e^{c_2/\lambda T}} \qquad (2)

The basic elements of this experimental question are sketched in Fig. F-2.

  • The emitter body is the tungsten filament of an incandescent lamp AA that emits a wide range of λ\lambda’s, and whose luminosity can be varied.
  • The test tube BB contains a liquid filter that only transmits a thin band of the visible spectrum around a value λ0\lambda_0 (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.
Sketch of the setup: lamp A fed by a battery through a variable resistor, emitting light towards a vertical test tube B containing a liquid; the transmitted light (arrow) reaches the photoresistor C on the right.

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

E∝uλ0R∝E−γ}⇒R∝uλ0−γ\left.\begin{aligned} E &\propto u_{\lambda_0} \\ R &\propto E^{-\gamma} \end{aligned}\right\} \Rightarrow \quad R \propto u_{\lambda_0}^{-\gamma}
The same family of u_λ versus λ curves with a narrow shaded vertical band marking the filter's transmission window around λ0, located on the rising (short-wavelength) side of the curves.

where the dimensionless parameter γ\gamma 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 RR and the filament temperature TT

R=c3ec2γ/λ0T(3)R = c_3 e^{c_2\gamma/\lambda_0 T} \qquad (3)

that we will use in page 6. In this relation c3c_3 is an unknown proportionality constant. By measuring RR as a function on TT one can obtain c2c_2, 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.

Drawing of the apparatus with numbered components: 1 platform with disk holding the LDR, tube and lamp supports and sockets marked Ω, V, A; 2 protecting cover (cylinder); 3 potentiometer; 4 battery with − and + terminals; 5 wires between platform and potentiometer; 6 wires between potentiometer and battery; 7, 8, 9 multimeters connected to the Ω, V and A sockets; 10 test tube; 11 stand for the test tube; 12 grey filter; 13 ruler.

EQUIPMENT:

  1. 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.
  2. Protecting cover.
  3. 10 turns and 1 kΩ potentiometer.
  4. 12 V battery.
  5. Red and black wires with plugs at both ends to connect platform to potentiometer.
  6. Red and black wires with plugs at one end and sockets for the battery at the other end.
  7. Multimeter to work as ohmmeter.
  8. Multimeter to work as voltmeter.
  9. Multimeter to work as ammeter.
  10. Test tube with liquid filter.
  11. Stand for the test tube.
  12. Grey filter.
  13. 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:
Five numbered set-up steps: 1 turning the potentiometer knob; 2 rotating the test-tube support between the lamp and LDR supports; 3 the LDR pushed towards the lateral hole of the tube support, opposite the lamp; 4 inserting the test tube into its support; 5 lowering the cylindrical cover onto the platform.
  1. Turn the potentiometer knob anticlockwise until reaching the end.
  2. 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.
  3. 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.
  4. Insert the test tube into its support.
  5. 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 RBR_B of a conducting filament can be given as

RB=ρlS(4)R_B = \rho\frac{l}{S} \qquad (4)

where ρ\rho is the resistivity of the conductor, ll is the length and SS 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,
T=3.05⋅108 ρ0.83(5)T = 3.05\cdot 10^{8}\,\rho^{0.83} \qquad (5)
  • 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

T=a RB0.83(6)T = a\,R_B^{0.83} \qquad (6)
  • Therefore, to get TT it is necessary to determine aa. This can be achieved by measuring the filament resistance RB,0R_{B,0} at ambient temperature T0T_0.

a) Measure with the multimeter the ambient temperature T0T_0. [1 т.]

b) It is not a good idea to use the ohmmeter to measure the filament resistance RB,0R_{B,0} at T0T_0 because it introduces a small unknown current that increases the filament temperature. Instead, to find RB,0R_{B,0} 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 RBR_B for each pair of values of VV and II, 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 RBR_B in the vertical axis against II. [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, RB,0R_{B,0}. Write the selected values in the Table for Task 2, c) in the Answer Sheets. Finally, obtain RB,0R_{B,0} and ΔRB,0\Delta R_{B,0}. [2,5 т.]

d) Compute the numerical values of aa and Δa\Delta a for RB,0R_{B,0} in Ω and T0T_0 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.

Graph of % transmittance (0 to 30) versus λ /nm (450 to 750): the curve is near zero up to about 540 nm, rises to a single peak of about 25 % near 590 nm and falls back to about zero near 670 nm.

Task 3 Determine λ0\lambda_0 and Δλ\Delta\lambda from Fig. F-6. [1 т.]

Note: 2 Δλ2\,\Delta\lambda is the total width at half height and λ0\lambda_0 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

R=bE−γ(7)R = bE^{-\gamma} \qquad (7)

where bb is a constant that depends on the composition and geometry of the LDR and γ\gamma is a dimensionless parameter that measures the variation of the resistance with the illumination EE produced by the incident radiation. Theoretically, an ideal LDR would have γ=1\gamma = 1, however many factors intervene, so that in the real case γ<1\gamma < 1.

Test tube between the lamp (right) and the LDR (left); light passes from the lamp through the tube to the LDR, shown by an arrow.

It is necessary to determine γ\gamma. This is achieved by measuring a pair RR and EE (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 E′=0.51 EE' = 0.51\ E. After measuring the resistance R′R' corresponding to this illumination, we have

R=bE−γ;R′=b(0.512 E)−γR = bE^{-\gamma} \quad ; \qquad R' = b\left(0.512\,E\right)^{-\gamma}
The same arrangement with a grey filter plate labelled F inserted between the lamp and the test tube.

From this

ln⁡RR′=γln⁡0.512(8)\ln\frac{R}{R'} = \gamma\ln 0.512 \qquad (8)

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 VV and II for VV in the range between 9.50 V and 11.50 V, and obtain the corresponding LDR resistances RR. (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>9.5V > 9.5 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 RR, 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 R′R'. Using these data in (8) compute γ\gamma and Δγ\Delta\gamma.

Perspective sketch of the test-tube support between the lamp and LDR holders (dotted) with the L-shaped grey filter being slid in between the lamp and the tube, indicated by an arrow.

[1,5 т.]

c) Modify Eq. (3) to display a linear dependence of ln RR on RB−0.83R_B^{-0.83}. 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 c2=hc/kc_2 = hc/k, compute hh and Δh\Delta h by any method (you are allowed to use statistical functions of the calculators provided by the organization). [3 т.]

(Speed of light, c=2.998⋅108 m s−1c = 2.998\cdot 10^{8}\ \mathrm{m\ s^{-1}} ; Boltzmann constant, k=1.381⋅10−23 J K−1k = 1.381\cdot 10^{-23}\ \mathrm{J\ K^{-1}})

Answer sheet 1

Draw the electric connections in the boxes and between boxes below.

Answer-sheet boxes to be connected: platform P with the photoresistor and bulb symbols, a red and a black socket on its right and the ohmmeter Ω, voltmeter V and ammeter A circles each with a red and a black socket on its left; potentiometer box Pm with the potentiometer symbol and two red and two black sockets; battery box B with a black and a red socket. Legend tables: Photoresistor, Incandescent Bulb, Potentiometer, Red socket, Black socket symbols; Ω Ohmmeter, V Voltmeter, A Ammeter, P Platform, Pm Potentiometer, B Battery.
Answer sheet 1

Answer sheet 2

TASK 2

a) (1.0 points)

T0=T_0=

b) (2.0 points)

VV, II, RBR_B

Vmin=V_{min}= *

  • This is a characteristic of your apparatus. You can’t go below it.
Complete original blank answer sheet 2, including printed task marks, table headings, empty response boxes and candidate identification fields. No answers or omitted column labels supplied.

Answer sheet 3

TASK 2

c) (2.5 points)

VV, II, RBR_B

RB0=R_{B0}=, ΔRB0=\Delta R_{B0}=

d) (1.0 points)

a=a=, Δa=\Delta a=

TASK 3 (1.0 points)

λ0=\lambda_0=, Δλ=\Delta\lambda=

Complete original blank answer sheet 3, including printed task marks, table headings, empty response boxes and candidate identification fields. No answers or omitted column labels supplied.

Answer sheet 4

TASK 4

a) (2.0 points)

VV, II, RR

b) (1.5 points)

R=R=, γ=\gamma=, R’=R’=, Δγ=\Delta\gamma=

c) (1.0 points)

Eq. (9)

Complete original blank answer sheet 4, including printed task marks, table headings, empty response boxes and candidate identification fields. No answers or omitted column labels supplied.

Answer sheet 5

TASK 4

d) (3.0 points)

VV, II, RR

e) (3.0 points)

h=h=, Δh=\Delta h=

Complete original blank answer sheet 5, including printed task marks, table headings, empty response boxes and candidate identification fields. No answers or omitted column labels supplied.

Отговори

Решение


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