Съдържание

Условие 1. Adjustments and Measurements: (5,3 т.)2. Calculations: (6,2 т.)3. Data analysis: (8,5 т.)Description of the Apparatus Отговори Решение

IPhO 2007, Експеримент — Задача 1. Determination of energy band gap of semiconductor thin films

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Всички материали от IPhO 2007

38th International Physics Olympiad, Isfahan-Iran, July 2007 — Experimental Problem

Внимание: Бележка към темата

The enclosed Description of the Apparatus is included from the companion source. The separate blank answer sheets for Tables 1d, 1e, 2c, 2d, 3a and 3d are absent from the available question, answer and apparatus packets. The two supplied answer editions contain the same official solution; it is presented once. Printed wording, repeated task labels and solution inconsistencies are preserved.

Условие

I. Introduction

Semiconductors can be roughly characterized as materials whose electronic properties fall somewhere between those of conductors and insulators. To understand semiconductor electronic properties, one can start with the photoelectric effect as a well-known phenomenon. The photoelectric effect is a quantum electronic phenomenon, in which photoelectrons are emitted from the matter through the absorption of sufficient energy from electromagnetic radiation (i.e. photons). The minimum energy which is required for the emission of an electron from a metal by light irradiation (photoelectron) is defined as "work function". Thus, only photons with a frequency $\nu$ higher than a characteristic threshold, i.e. with an energy $h\nu$ ( $h$ is the Planck’s constant) more than the material’s work function, are able to knock out the photoelectrons.

Three wavy red arrows (photons) strike a metal plate containing electrons drawn as small circles with minus signs; two straight red arrows show electrons ejected upward from the surface. — Figure 1. An illustration of photoelectron emission from a metal plate: The incoming photon should have an energy which is more than the work function of the material.

In fact, the concept of work function in the photoelectric process is similar to the concept of the energy band gap of a semiconducting material. In solid state physics, the band gap $E_g$ is the energy difference between the top of the valence band and the bottom of the conduction band of insulators and semiconductors. The valence band is completely filled with electrons, while the conduction band is empty however electrons can go from the valence band to the conduction band if they acquire sufficient energy (at least equal to the band gap energy).The semiconductor's conductivity strongly depends on its energy band gap.

Energy band diagram with a vertical arrow labelled E. A hatched unfilled band at the top labelled 'Unfilled band' and 'Conduction band'; below a gap marked 'Energy Band Gap'; a grey band labelled 'Valence band' and below it further grey bands labelled 'Filled bands', all containing electrons. Wavy red arrows (photons) and straight black arrows show electrons excited from the valence band to the conduction band. — Figure 2. Energy band scheme for a semiconductor.

Band gap engineering is the process of controlling or altering the band gap of a material by controlling the composition of certain semiconductor alloys. Recently, it has been shown that by changing the nanostructure of a semiconductor it is possible to manipulate its band gap.

In this experiment, we are going to obtain the energy band gap of a thin-film semiconductor containing nano-particle chains of iron oxide (Fe $_2$ O $_3$ ) by using an optical method. To measure the band gap, we study the optical absorption properties of the transparent film using its optical transmission spectrum. As a rough statement, the absorption spectra shows a sharp increase when the energy of the incident photons equals to the energy band gap.

II. Experimental Setup

You will find the following items on your desk:

A large white box containing a spectrometer with a halogen lamp.
A small box containing a sample, a glass substrate, a sample-holder, a grating, and a photoresistor.
A multimeter.
A calculator.
A ruler.
A card with a hole punched in its center.
A set of blank labels.

The spectrometer contains a goniometer with a precision of $5'$ . The Halogen lamp acts as the source of radiation and is installed onto the fixed arm of the spectrometer (for detailed information see the enclosed "Description of Apparatus").

The small box contains the following items:

A sample-holder with two windows: a glass substrate coated with Fe $_2$ O $_3$ film mounted on one window and an uncoated glass substrate mounted on the other.
A photoresistor mounted on its holder, which acts as a light detector.
A transparent diffraction grating (600 line/mm).

Note: Avoid touching the surface of any component in the small box!

A schematic diagram of the setup is shown in Figure 3:

Schematic of the setup: Halogen lamp on the right shining through Diffusive glass and a Lens onto a Grating (600 lines/mm) at the centre of a Goniometer; the Optical axis is marked, the rotatable arm at angle θ carries the Sample in front of the Entrance hole and a Photoresistor at its end, connected by wires to an Ohmmeter (Max. range 200 MΩ). — Figure 3. Schematic diagram of the experimental setup.

III. Methods

To obtain the transmission of a film at each wavelength, $T_{film}(\lambda)$ , one can use the following formula:

T_{film}(\lambda)=I_{film}(\lambda)/I_{glass}(\lambda)\qquad(1)

where $I_{film}$ and $I_{glass}$ are respectively the intensity of the light transmitted from the coated glass substrate, and the intensity of the light transmitted from the uncoated glass slide. The value of $I$ can be measured using a light detector such as a photoresistor. In a photoresistor, the electrical resistance decreases when the intensity of the incident light increases. Here, the value of $I$ can be determined from the following relation:

I(\lambda)=C(\lambda)R^{-1}\qquad\mathbf{(2)}

where $R$ is the electrical resistance of the photoresistor, $C$ is a $\lambda$ -dependent coefficient.

The transparent grating on the spectrometer diffracts different wavelengths of light into different angles. Therefore, to study the variations of $T$ as a function of $\lambda$ , it is enough to change the angle of the photoresistor ( $\theta'$ ) with respect to the optical axis (defined as the direction of the incident light beam on the grating), as shown in Figure 4.

From the principal equation of a diffraction grating:

n\lambda=d\,[\sin(\theta'-\theta_0)+\sin\theta_0]\qquad(3)

one can obtain the angle $\theta'$ corresponding to a particular $\lambda$ : $n$ is an integer number representing the order of diffraction, $d$ is the period of the grating, and $\theta_o$ is the angle the normal vector to the surface of grating makes with the optical axis (see Fig. 4). (In this experiment we shall try to place the grating perpendicular to the optical axis making $\theta_o=0$ , but since this cannot be achieved with perfect precision the error associated with this adjustment will be measured in task 1-e.)

A dotted line labelled Grating crosses a horizontal arrow labelled Optical axis; the grating normal makes angle θo with the optical axis (marked twice, near the normal and near the grating), and a diffracted ray makes angle θ' with the optical axis. — Figure 4. Definition of the angles involved in Equation 3.

Experimentally it has been shown that for photon energies slightly larger than the band gap energy, the following relation holds:

\alpha h\nu=A(h\nu-E_g)^{\eta}\qquad(4)

where $\alpha$ is the absorption coefficient of the film, $A$ is a constant that depends on the film’s material, and $\eta$ is the constant determined by the absorption mechanism of the film’s material and structure. Transmission is related to the value of $\alpha$ through the well-known absorption relation:

T_{film}=\exp(-\alpha\,t)\qquad(5)

where $t$ is thickness of the film.

IV. Tasks:

0. Your apparatus and sample box (small box containing the sample holder) are marked with numbers. Write down the Apparatus number and Sample number in their appropriate boxes, in the answer sheet.

1. Adjustments and Measurements: (5,3 т.)

1-a Check the vernier scale and report the maximum precision ( $\Delta\theta$ ). [0,1 т.]

Note: Magnifying glasses are available on request.

Step1:

To start the experiment, turn on the Halogen lamp to warm up. It would be better not to turn off the lamp during the experiment. Since the halogen lamp heats up during the experiment, please be careful not to touch it.

Place the lamp as far from the lens as possible, this will give you a parallel light beam.

We are going to make a rough zero-adjustment of the goniometer without utilizing the photoresistor. Unlock the rotatable arm with screw 18 (underneath the arm), and visually align the rotatable arm with the optical axis. Now, firmly lock the rotatable arm with screw 18. Unlock the vernier with screw 9 and rotate the stage to 0 on the vernier scale. Now firmly lock the vernier with screw 9 and use the vernier fine-adjustment screw (screw 10) to set the zero of the vernier scale. Place the grating inside its holder. Rotate the grating's stage until the diffraction grating is roughly perpendicular to the optical axis. Place the card with a hole in front of the light source and position the hole such that a beam of light is incident on the grating. Carefully rotate the grating so that the spot of reflected light falls onto the hole. Then the reflected light beam coincides with the incident beam. Now lock the grating's stage by tightening screw 12.

1-b By measuring the distance between the hole and the grating, estimate the precision of this adjustment ( $\Delta\theta_o$ ). 0.3 pt

Now, by rotating the rotatable arm, determine and report the range of angles for which the first-order diffraction of visible light (from blue to red) is observed. 0.2 pt [0,5 т.]

Step 2:

Now, install the photoresistor at the end of the rotatable arm. To align the system optically, by using the photoresistor, loosen the screw 18, and slightly turn the rotatable arm so that the photoresistor shows a minimum resistance. For fine positioning, firmly lock screw 18, and use the fine adjustment screw of the rotatable arm.

Use the vernier fine-adjustment screw to set the zero of the vernier scale.

1-c Report the measured minimum resistance value ( $R^{(0)}_{\min}$ ). 0.1 pt

Your zero-adjustment is more accurate now, report the precision of this new adjustment ( $\Delta\varphi_o$ ).

Note: $\Delta\varphi_o$ is the error in this alignment i.e. it is a measure of misalignment of the rotatable arm and the optical axis. 0.1 pt

Hint: After this task you should tighten the fixing screws of the vernier. Moreover, tighten the screw of the photoresistor holder to fix it and do not remove it during the experiment.

Step 3:

Move the rotatable arm to the region of the first-order diffraction. Find the angle at which the resistance of the photoresistor is minimum (maximum light intensity). Using the balancing screws, you can slightly change the tilt of the grating’s stage, to achieve an even lower resistance value.

Report the minimum value of the observed resistance ( $R^{(1)}_{\min}$ ) in its appropriate box. 0.1 pt [0,3 т.]

It is now necessary to check the perpendicularity of the grating for zero adjustment, again. For this you must use the reflection-coincidence method of Step 1.

Important: From here onwards carry out the experiment in dark (close the cover).

Measurements: Screw the sample-holder onto the rotatable arm. Before you start the measurements, examine the appearance of your semiconductor film (sample). Place the sample in front of the entrance hole $S_1$ on the rotatable arm such that a uniformly coated part of the sample covers the hole. To make sure that every time you will be working with the same part of the sample make proper markings on the sample holder and the rotatable arm with blank labels.

Attention: At higher resistance measurements it is necessary to allow the photoresistor to relax, therefore for each measurement in this range wait 3 to 4 minutes before recording your measurement.

1-d Measure the resistance of the photoresistor for the uncoated glass substrate and the glass substrate coated with semiconductor layer as a function of the angle $\theta$ (the value read by the goniometer for the angle between the photoresistor and your specified optical axis). Then fill in Table 1d. Note that you need at least 20 data points in the range you found in Step 1b. Carry out your measurement using the appropriate range of your ohmmeter. 2.0 pt

Consider the error associated with each data point. Base your answer only on your direct readings of the ohmmeter. 1.0 pt [3 т.]

Step 4:

The precision obtained so far is still limited since it is impossible to align the rotatable arm with the optical axis and/or position the grating perpendicular to the optical axis with 100% precision. So we still need to find the asymmetry of the measured transmission at both sides of the optical axis (resulting from the deviation of the normal to the grating surface from the optical axis ( $\theta_o$ )).

To measure this asymmetry, follow these steps:

1-e First, measure $T_{film}$ at $\theta=-20^{\circ}$ . Then, obtain values for $T_{film}$ at some other angles around $+20^{\circ}$ . Complete Table 1e (you can use the values obtained in Table 1d). 0.6 pt

Draw $T_{film}$ versus $\theta$ and visually draw a curve. 0.6 pt

On your curve find the angle $\gamma$ for which the value of $T_{film}$ is equal to the $T_{film}$ that you measured at $\theta=-20^{o}$ ( $\gamma\equiv\theta\big|_{T_{film}=T_{film}(-20^{\circ})}$ ). Denote the difference of this angle with $+20^{\circ}$ as $\delta$ , in other words:

\delta=\gamma-20^{\circ}\qquad(6)

Report the value of $\delta$ in the specified box. 0.2 pt [1,4 т.]

Then for the first-order diffraction, Eq. (3) can be simplified as follows:

\lambda=d\sin(\theta-\delta/2)\,,\qquad(7)

where $\theta$ is the angle read on the goniometer.

2. Calculations: (6,2 т.)

2-a Use Eq. (7) to express $\Delta\lambda$ in terms of the errors of the other parameters (assume $d$ is exact and there is no error is associated with it). Also using Eqs. (1), (2), and (5), express $\Delta T_{film}$ in terms of $R$ and $\Delta R$ . [0,6 т.]

2-b Report the range of values of $\Delta\lambda$ over the region of first-order diffraction. [0,3 т.]

2-c Based on the measured parameters in Task 1, complete Table 2c for each $\theta$ . Note that the wavelength should be calculated using Eq. (7). [2,4 т.]

2-d Plot $R^{-1}_{glass}$ and $R^{-1}_{film}$ as a function of wavelength together on the same diagram. Note that on the basis of Eq. (2) behaviors of $R^{-1}_{glass}$ and $R^{-1}_{film}$ can reasonably give us an indication of the way $I_{glass}$ and $I_{film}$ behave, respectively. 1.5 pt

In Table 2d, report the wavelengths at which $R_{glass}$ and $R_{film}$ attain their minimum values. 0.4 pt [1,9 т.]

2-e For the semiconductor layer (sample) plot $T_{film}$ as a function of wavelength. This quantity also represents the variation of the film transmission in terms of wavelength. [1 т.]

3. Data analysis: (8,5 т.)

By substituting $\eta=1/2$ and $A=0.071$ $((\mathrm{eV})^{1/2}/\mathrm{nm})$ in Eq. (4) one can find values for $E_g$ and $t$ in units of eV and nm, respectively. This will be accomplished by plotting a suitable diagram in an $x-y$ coordinate system and doing an extrapolation in the region satisfying this equation.

3-a By assuming $x=h\nu$ and $y=(\alpha\,t\,h\nu)^2$ and by using your measurements in Task 1, fill in Table 3a for wavelengths around 530 nm and higher. Express your results ( $x$ and $y$ ) with the correct number of significant figures (digits), based on the estimation of the error on one single data point.

Note that $h\nu$ should be calculated in units of eV and wavelength in units of nm. Write the unit of each variable between the parentheses in the top row of the table. [2,4 т.]

3-b Plot $y$ versus $x$ .

Note that the $y$ parameter corresponds to the absorption of the film. Fit a line to the points in the linear region around 530 nm.

Specify the region where Eq. (4) is satisfied, by reporting the values of the smallest and the largest x-coordinates for the data points to which you fit the line. [2,6 т.]

3-c Call the slope of this line $m$ , and find an expression for the film thickness ( $t$ ) and its error ( $\Delta t$ ) in terms of $m$ and $A$ (consider $A$ to have no error). [0,5 т.]

3-d Obtain the values of $E_g$ and $t$ and their associated errors in units of eV and nm, respectively. Fill in Table 3d. [3 т.]

Some useful physical constants required for your analysis:

Speed of the light: $c=3.00\times10^{8}\ \mathrm{m/s}$
Plank’s constant: $h=6.63\times10^{-34}\ \mathrm{J.s}$
Electron charge: $e=1.60\times10^{-19}\ \mathrm{C}$

Description of the Apparatus

In Fig.1 you can see the general view of the apparatus set up on your desk, which will be used in the experiment. The instrument is a spectroscope to be equipped with a detector to act as a simple spectrometer.

To start adjusting the apparatus, you should first pull up the white cover of the box (Fig.1). The cover pivots on one side of the base of the apparatus. In order to establish a dark environment for the detector, the cover should be returned to its initial position and kept tightly closed during the measurement of the spectra. The power cord has a switch that turns the halogen lamp on and off. There are four screws to level the apparatus (a magnified view of which you can see in right inset of Fig.1)

Open white cover over the spectroscope. A level-adjusting screw on the base is enlarged in the right inset. — Figure 1. Apparatus of the experiment. One of the level adjusting screws is enlarged in the right inset.

Warning 1: Avoid touching the halogen lamp and its holder which will be hot after the lamp is turned on!

Warning 2: Do not manipulate the adaptor and its connections. Power is supplied to the apparatus through 220 V outlets!

The top view of the apparatus is shown in Fig.2 . The details are introduced in the figure.

Top view of the spectroscope apparatus with numbered components 1–24 and apparatus number 312 in its top-left corner. — Figure 2.

Power cord
Halogen lamp and its cooling fan
On/Off switch
Arm of adjustable length
Adjusting screw
Adaptor: 220V – less than 12 V
Lens
Vernier
Vernier’s lock
Fine adjustment screw for the vernier
Grating’s stage
Grating’s stage’s fixing screw
Adjustment screw for leveling the grating’s stage (shown in Fig. 4)
Grating holder
Sample holder
Fixing and adjusting screw for the sample and glass holder (Fig. 6)
Rotatable arm
Rotatable arm’s lock (Fig.4 )
Fine adjustment for the rotatable arm
Detector position
Fixing screw for the detector
Connecting socket for the detector
Connection to the multimeter
Fixing screw to the base

The number mentioned on the top-left corner, is the apparatus number.

The angle, which the rotatable arm makes with the direction of the fixed arm of the apparatus, could be measured by a protractor equipped by a vernier. In this vernier resolution scale is $30'$ (minutes of arc). This instrument is able to measure an angle with accuracy of $5'$ .

In addition to the apparatus you should find a box (Figure 3), containing the following elements:

1: a detector in its holder; 2: a 600 line/mm grating; 3: the sample and a glass substrate mounted in a frame.

Small box containing sample and glass holder (3), diffraction grating (2), and photoresistor in its holder (1). — Figure 3. The small box, containing the glass and sample holder, a diffraction grating and a photoresistor.

First, you should take the grating out of its cover and put it into its frame (the grating holder, Fig. 4), carefully.

CAUTION: Touching the surface of the grating could reduce its diffraction efficiency seriously, or even damage it!

There are three adjustment screws (Fig. 4) for making the grating stand vertically in its position.

Three views of grating and stage fixing and adjustment screws. Labels A1, A2, 1, 7, 9, 10, 12, 13 and 18 are retained. — Figure 4. Locking, fixing and adjusting screws of the apparatus. $A_1$ : Fixing screw for the grating; $A_2$ : The grating. 7, 9, 10, 12-14, 18 and 19 are explained in Figure 2.

The detector should be tight to its position, in the end of the rotatable arm, (Figure 5):

Front and mounted views of the detector with D1 photoresistor, D2 connecting wire, D3 holder, and apparatus labels 17 and 21. — Figure 5 . The detector and its holder. D1: The photoresistor; D2: connecting wire. D3: The detector holder. 17 and 21 are explained in Fig. 2.

The sample and the glass substrate are fixed to a frame (holder) (Fig. 6c), which would be attached to the instrument by a fixing screw (Fig. 6a, item 16). This frame is rotatable and one can put the sample or the glass substrate in front of the entrance hole, by turning the frame around the fixing screw (Fig. 6a).

Three panels a, b, c of sample and glass holder, with entrance hole S1, sample S2, glass substrate S3, and apparatus labels A2, 15 and 16. — Figure 6 . The Sample and the glass holder. S1: Entrance hole; S2: Sample; S3: Glass substrate. 15 and 16 are explained in Fig. 2.

The Multimeter which you should use for recording the signal detected by the photoresistor is shown in the Fig. 7. This multimeter can measure up to $200\ \mathrm{M\Omega}$ . The red and black probe wires should be connected to the instrument as is shown in the Fig. 7. The on/off button is placed on the left hand side of the multimeter (Fig. 7, item M1).

Digital multimeter and probe wires. Labels M1 on/off switch, M2 probe wires, M3 Hold button and M4 apparatus connections. — Figure 7. The Multimeter for measuring the resistance of the photoresistor. M1: on/off switch; M2: probe wires; M3: Hold button; M4: connections to the apparatus.

Note: The multimeter has auto-off feature. In the case of auto-off, you should push on/off button (M1) twice, successively.

Hold button should not be active during the experiment.

Отговори

Решение

Solution (The Experimental Question):

Task 1

1a. $\Delta\theta_{\mathrm{nominal}}=5'=0.08^{\circ}$

$\Delta\theta_{\mathrm{nominal}}$ (degree)	0.08

1b.

Diagram for 1b: a dotted grating line tilted from the vertical, a horizontal dashed optical axis; the incident beam travels along the axis from a hole in a vertical card at distance a from the grating, and the reflected beam returns at angle θ0 to the axis, hitting the card at distance r above the hole. Labels: Reflected Beam, Incident Beam, Optical Axis, a, r, θ0.

If “a” is the distance between card and the grating and “r” is the distance between the hole and the light spot so we have

\Delta f(x_1,x_2,\ldots)=\sqrt{\left(\frac{\partial f}{\partial x_1}\Delta x_1\right)^2+\left(\frac{\partial f}{\partial x_2}\Delta x_2\right)^2+\ldots}

\tan(2\theta_0)=\frac{r}{a},\ \mathit{If}\ \theta_0\ll 1\Rightarrow\theta_0=\frac{r}{2a}\Rightarrow\Delta\theta_0=\sqrt{\left(\frac{\Delta r}{2a}\right)^2+\left(\frac{r\,\Delta a}{2a^2}\right)^2}

We want $\theta_0$ to be zero i.e. $r=0\Rightarrow\Delta\theta_0=\frac{\Delta r}{2a}$

\Delta r=1\,\mathrm{mm},\ a=(70\pm1)\,\mathrm{mm}\Rightarrow\theta_0=\frac{\Delta r}{2a}\ \mathrm{rad}=0.007\ \mathrm{rad}=0.4^{\circ}

$\Delta\theta_0$	$0.4^{\circ}$
$\theta$ range of visible light (degree)	$13^{\circ}\le\theta\le 26^{\circ}$

1c.

$R^{(0)}_{\min}$	(21.6±0.1) kΩ
$\Delta\varphi_0$	$5' = 0.08^{\circ}$
$R^{(1)}_{\min}$	R=(192±1) kΩ

$\Delta\varphi_0=5'$ because

$\theta=5'\Rightarrow R=(21.9\pm0.1)\ \mathrm{k\Omega}$

$\theta=-5'\Rightarrow R=(21.9\pm0.1)\ \mathrm{k\Omega}$

1d. Table 1d. The measured parameters

θ (degree)	$R_{\mathrm{glass}}$ (MΩ)	$\Delta R_{\mathrm{glass}}$ (MΩ)	$R_{\mathrm{film}}$ (MΩ)	$\Delta R_{\mathrm{film}}$ (MΩ)
15.00	3.77	0.03	183	3
15.50	2.58	0.02	132	2
16.00	1.88	0.01	87	1
16.50	1.19	0.01	51.5	0.5
17.00	0.89	0.01	33.4	0.3
17.50	0.68	0.01	19.4	0.1
18.00	0.486	0.005	10.4	0.1
18.50	0.365	0.005	5.40	0.03
19.00	0.274	0.003	2.66	0.02
19.50	0.225	0.002	1.42	0.01
20.00	0.200	0.002	0.880	0.005
20.50	0.227	0.002	0.822	0.005
21.00	0.368	0.003	1.123	0.007
21.50	0.600	0.005	1.61	0.01
22.00	0.775	0.005	1.85	0.01
22.50	0.83	0.01	1.87	0.01
23.00	0.88	0.01	1.93	0.02
23.50	1.01	0.01	2.14	0.02
24.00	1.21	0.01	2.58	0.02
24.50	1.54	0.01	3.27	0.02
25.00	1.91	0.01	4.13	0.02
16.25	1.38	0.01	66.5	0.5
16.75	1.00	0.01	40.0	0.3
17.25	0.72	0.01	23.4	0.2
17.75	0.535	0.005	12.8	0.1
18.25	0.391	0.003	6.83	0.05
18.75	0.293	0.003	3.46	0.02
19.25	0.235	0.003	1.76	0.01
19.75	0.195	0.002	0.988	0.005
20.25	0.201	0.002	0.776	0.005
20.75	0.273	0.003	0.89	0.01

1e. In θ=-20˚ $\Rightarrow$ $R_{\mathrm{glass}}$ = (132± 2) kΩ , $R_{\mathrm{film}}$ = (518±5) kΩ

$\theta$	$T_{\mathrm{film}}$	$\theta$	$T_{\mathrm{film}}$
$\theta = -20^{\circ}$	0.255	19.25	0.134
		19.50	0.158
		19.75	0.197
		20.00	0.227
		20.25	0.259
		20.50	0.276
		20.75	0.307

Graphics

Graph for 1e: Transmission (0.10 to 0.35) versus θ (degree) (19.0 to 21.0); seven data points rising from about 0.134 at 19.25° to 0.307 at 20.75° with a smooth curve through them, and a horizontal dashed line at 0.255 labelled T_film|θ=-20°.

We see that: $T(\theta=20.25^{\circ})=T(\theta=-20^{\circ})$

$\delta$ (degree)	0.25±0.08

Task 2.

2a.

\lambda=d\sin\left(\theta-\frac{\delta}{2}\right)\Rightarrow\Delta\lambda=\lambda\sqrt{\left(\frac{\Delta d}{d}\right)^2+\cot^2\left(\theta-\frac{\delta}{2}\right)\left(\Delta\theta^2+\frac{\Delta\delta^2}{4}\right)}\approx d\ \cos(\theta)\left(\frac{0.1\pi}{180}\right)

where $\Delta\theta=\Delta\delta=5'=0.08$ degree

and $d=\frac{1}{600}$ mm

\boxed{\Delta\lambda=2.9\cos(\theta)\ (\mathrm{nm})}

T_{film}=\frac{R_{glass}}{R_{film}}\Rightarrow\Delta T=T_{film}\sqrt{\left(\frac{\Delta R_{film}}{R_{film}}\right)^2+\left(\frac{\Delta R_{glass}}{R_{glass}}\right)^2}

\boxed{\Delta T=\frac{R_{glass}}{R_{film}}\sqrt{\left(\frac{\Delta R_{film}}{R_{film}}\right)^2+\left(\frac{\Delta R_{glass}}{R_{glass}}\right)^2}}

2b.

13\le\theta\le 26

\boxed{2.6\le\Delta\lambda\le 2.8\ \mathrm{nm}}

2c. Table 2c. The calculated parameters using the measured parameters

$\theta$ (degree)	λ (nm)	$I_g/C(\lambda)$ ( $\mathrm{M\Omega^{-1}}$ )	$I_s/C(\lambda)$ ( $\mathrm{M\Omega^{-1}}$ )	$T_{\mathrm{film}}$	αt
15.0	428	0.265	0.00546	0.0206	3.88
15.5	442	0.388	0.00758	0.0195	3.94
16.0	456	0.532	0.0115	0.0216	3.83
16.25	463	0.725	0.0150	0.0208	3.88
16.5	470	0.840	0.0194	0.0231	3.77
16.75	477	1.00	0.0250	0.0250	3.69
17.0	484	1.12	0.0299	0.0266	3.63
17.25	491	1.39	0.0427	0.0308	3.48
17.5	498	1.47	0.0515	0.0351	3.35
17.75	505	1.87	0.0781	0.0418	3.17
18.0	512	2.06	0.096	0.0467	3.06
18.25	518	2.56	0.146	0.0572	2.86
18.5	525	2.74	0.185	0.0676	2.69
18.75	532	3.41	0.289	0.0847	2.47
19.0	539	3.65	0.376	0.103	2.27
19.25	546	4.26	0.568	0.134	2.01
19.5	553	4.44	0.704	0.158	1.84
19.75	560	5.13	1.01	0.197	1.62
20.0	567	5.00	1.14	0.227	1.48
20.25	573	4.98	1.29	0.259	1.35
20.5	580	4.41	1.22	0.276	1.29
20.75	587	3.66	1.12	0.307	1.18
21.0	594	2.72	0.890	0.328	1.12
21.5	607	1.67	0.621	0.373	0.99
22.0	621	1.29	0.541	0.419	0.87
22.5	634	1.20	0.535	0.444	0.81
23.0	648	1.14	0.518	0.456	0.79
23.5	661	0.99	0.467	0.472	0.75
24.0	675	0.826	0.388	0.469	0.76
24.5	688	0.649	0.306	0.471	0.75
25.0	701	0.524	0.242	0.462	0.77

2d. Graphics

Graph for 2d: R^-1 (MΩ^-1) from 0.0 to 6.0 versus Wavelength (nm) from 420 to 720; dark blue diamonds labelled 'glass' rise to a peak of about 5.1 near 560 nm and fall again; pink squares labelled 'film' stay near zero below 500 nm, peak at about 1.3 near 573 nm and decrease slowly.

$\lambda_{\mathrm{max}}(I_{\mathrm{glass}})$	564±5 (nm)
$\lambda_{\mathrm{max}}(I_{\mathrm{film}})$	573±5 (nm)

2e. Graphics

Graph for 2e: Transmission (0.00 to 0.50) versus Wavelength(nm) (420 to 720); data points joined by a line stay near 0.02 up to about 500 nm, rise steeply between 520 and 620 nm and level off at about 0.47 above 660 nm.

Task 3.

3a. Table 3a. The calculated parameters for each measured data point

$\theta$ (degree)	$x$ (eV)	$y$ ( eV $^2$ )
15.00	2.898	126.6
15.50	2.806	121.9
16.00	2.720	108.8
16.25	2.679	107.8
16.50	2.639	98.9
16.75	2.600	92.0
17.00	2.563	86.3
17.25	2.527	77.4
17.50	2.491	69.7
17.75	2.457	60.9
18.00	2.424	55.1
18.25	2.392	46.8
18.50	2.360	40.4
18.75	2.330	33.1
19.00	2.300	27.3
19.25	2.271	20.91
19.50	2.243	17.07
19.75	2.215	12.92
20.00	2.188	10.51
20.25	2.162	8.53
20.50	2.137	7.56
20.75	2.112	6.23
21.00	2.088	5.43
21.50	2.041	4.06
22.00	1.997	3.02
22.50	1.954	2.52
23.00	1.914	2.26
23.50	1.875	1.98
24.00	1.838	1.94
24.50	1.803	1.84
25.00	1.769	1.86

3b. Graphics

Graph for 3b: (α t h ν)^2 (eV^2) from 0 to 140 versus hν (eV) from 1.7 to 3.0; data points with small error bars stay near zero below 2.1 eV, rise linearly between about 2.24 and 2.68 eV and flatten above; a dashed straight-line fit crosses the x-axis near 2.17 eV; two thick black arrows mark the ends of the fitted linear region.

$\mathrm{x}_{\min}=2.24(\mathrm{eV})$	$\mathrm{x}_{\max}=2.68(\mathrm{eV})$

3c.

\alpha h\nu=A\left(h\nu-E_g\right)^{\frac{1}{2}}\Rightarrow\left(\alpha t\,h\nu\right)^2=\left(A\,t\right)^2\left(h\nu-E_g\right)

\Rightarrow y=\left(A\,t\right)^2\left(x-E_g\right)\Rightarrow m=\left(A\,t\right)^2\Rightarrow t=\frac{\sqrt{m}}{A}

\Rightarrow\frac{\Delta t}{t}=\frac{\Delta m}{2m}

$t=\frac{\sqrt{m}}{A}$
$\Delta t=\frac{\Delta m}{2A\sqrt{m}}$

3d. In linear range we have, m=213 (eV), r $^2$ = 0.9986, E $_g$ =2.17 (eV)
and we have A = 0.071 $(\mathrm{eV}^{1/2}/\mathrm{nm})$ so we find t= 206 (nm)

\Delta m=\sqrt{\frac{(\delta y)^2+\frac{m^2}{R^2}(\delta x)^2}{\sum_i x_i^2-N\bar{x}^2}}\approx\sqrt{\frac{(\delta y)^2+(m\,\delta x)^2}{\sum_i x_i^2-N\bar{x}^2}}=\sqrt{\frac{(\delta xy)^2}{\sum_i x_i^2-N\bar{x}^2}},\ (\delta xy)^2=(\delta y)^2+(m\,\delta x)^2

where $\delta x$ & $\delta y$ are the mean of error range of x & y

\delta x\approx\sqrt{\frac{\sum_i\delta x_i^{\,2}}{N}}\ \&\ \delta y=\sqrt{\frac{\sum_i\delta y_i^{\,2}}{N}}\quad\mathrm{So}\quad\delta x\approx 0.014\ (\mathrm{eV}),\ \delta y\approx 0.9(\mathrm{eV})^2

$\rightarrow\Delta m\approx 10\ (\mathrm{eV})\rightarrow\Delta t=t\times\Delta m/(2\,m)\approx 5\ (\mathrm{nm})$

\Delta E_g=\frac{1}{m}\sqrt{\left(\left(\frac{m^2\,\delta x^2+\delta y^2}{N}\right)+\left(\frac{\bar{y}}{m}\right)^2\Delta m^2\right)}=\frac{1}{m}\sqrt{\left(\left(\frac{\delta xy^2}{N}\right)+\left(\frac{\bar{y}}{m}\right)^2\Delta m^2\right)}

\Delta E_g\approx 0.02(\mathrm{eV})

Table 3d. The calculated values of $E_g$ and $t$ using Fig. 3

$E_g$ (eV)	$\Delta E_g$ (eV)	$t$ (nm)	$\Delta t$ (nm)
2.17	0.02	206	5

Оригинал в Архива: IPhO_2007_Exp(Green)_question.pdf · официални решения: IPhO_2007_Exp(Green)_answer.pdf · допълнителни данни: exp.pdf · допълнителни данни: exp_sol.pdf