Lab Report 5 - Experiment 6: The Photoelectric Effect PDF

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Physics Experimental Report for 100 level...

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PH103 – Quantum & Electrical Physics Experiment 6: The Photoelectric Effect Aim: This experiment aims for the following [ CITATION Kum13 \l 3081 ];  To examine some aspects of the photoelectric effect.  also, to determine the Planck’s constant value denoted as ‘h’.

Apparatus: The following list below are the electronic components required to conduct the experiment [CITATION Kum13 \l 3081 ];

 IEC Photoelectric Unit which comprises a photo emissive cell,  a high gain transistor amplifier,  a calibrated voltage divider,  a dry battery of 9V,  the CRO Unit and the Color filters  and a tungsten filament lamp of 40W.

IEC Photoelectric Unit

Results Data Analysis:

There are three parts carried out in this experiment. The first part is called the Photo-electric Effect which it explains the effect of photo electric for the backing voltage and the tube current. The second part is called Planck’s Constant, h where it explains the purpose of backing voltage increasing until the tube current gives zero and it also will tabulate the data table provided in the manual together with graphs, calculation and comparison. And the third and last part is called relationship between Illumination and Current where it shows a data table together with a graph and observational reasons regarding the phototube current. Part A: Photo Electric Effect The photoelectric effect for the backing voltage was recorded as 0Volts whereas the tube current observed to be o.47µA. The reading of the tube current happens or occur the moment the light is connected to the rails of the tube. Hence, 0.47µA is the amount of current that has been passed due to the electrons being omitted from the metal surface in the photoelectric unit Part B: Planck’s Constant (h) What is the purpose of doing this? Step 3 The following table presented the measured and calculated data of all values needed for this particular part and will also be used to sketch the graph of backing voltage against frequency, obtaining the Planck constant and its percentage error. Also, below (above the table) is the sample calculation of the frequency and the average backing voltage in which filter color orange is used. Note that the speed of light, c is 3 ×108 m /s wavelength which is given in the table below. Sample calculation of Frequency & Backing Voltage:  Frequency (Orange), c 3 ×10 8 =5.6604 ×10 14 Hz f= = −9 λ 530 × 10  Average backing voltage (Orange),

|

V=

||

|

T 1+T 2 +T 3 (−0.590) + (−0.588 ) +(−0.589 ) = =0.589 3 3

whereas

λ

is

Note that all the average backing voltage for each filter color is negative, however we use the absolute value for each due to the graph of backing voltage versus frequency. Frequency Filter

Wavelength

(

Color

(nm)

×10 Hz

14

Backing Voltage (V)

Trial 1

Trial 2

Trial 3

Average

) Red

590

5.085

-0.339

-0.340

-0.338

0.339

Orange

530

5.660

-0.590

-0.588

-0.589

0.589

Yellow

492

6.098

-0.730

-0.735

-0.733

0.733

Green

460

6.522

-0.850

-0.890

-0.880

0.873

Blue

428

7.009

-1.041

-1.043

-1.040

1.041

From the above table, a graph of backing voltage against frequency is drawn as shown below. There are two graphs showcasing the exact same relations. The first one shows the graph at a distance lookout and at the same time if the trend line is extended till it intersects the x-axis and the y-axis, it can be observed that the y-intercept is negative so therefore the backing voltage is a negative value of around -1.4693V when the frequency is 0Hz. Whereas the second one shows the clear image of the relationship of backing voltage against the frequency.

Backing Voltage Vs Frequency 1.2 1

f(x) = 0.36 x − 1.47

0.8

Backing Voltage (V)

0.6 0.4 0.2 -2

-1

0

0

1

2

3

4

5

6

7

8

-0.2 -0.4 -0.6 -0.8 -1 -1.2

Frequency ((×10^14 ��)

Backing Voltage Vs Frequency 1.2

Backing Voltage (V)

1

f(x) = 0.36 x − 1.47

0.8

0.6

0.4

0.2 5

5.5

6

6.5

7

7.5

Frequency ((×10^14 ��)

According to the theory behind photoelectric effect, we know that the electro-voltage, eV is equal to the product of planck’s constant, h and frequency, f subtracting the work function,

φ. And when the electro-voltage, eV equation compares to the quadratic equation, there are some comparisons which can be determined and one is that the slope of the graph of backing voltage versus frequency is equals to the product of planck’s constant and frequency in electro-volts. Another comparison is that the work function, ɸ is equals to the y-intercept of the graph. The equation of the trend line of the graph above is approximately equals to 0.3596x – 1.4693. All the details expressions are shown below. Note that f =

c . λ

eV =hf −ϕ ≅ y=mx + c V=

ϕ m ×e hc hc ϕ hc 1 ϕ ∧ y−∫ .= − , so the slope , m= → h= − = c eλ e e λ e e e

¿ the graph above ,the trend line equationis y=0.3596 x −1.4693

thereforem=0.3596∧ y −∫ .=1.4693 So,

the

planck’s

constant

determine

the

graph

above

is

−28

J ∨1917866.667 ×10

−34

J .

−26

%=0.00 … 001918 % ≈ 0 % . Hence, in comparison of the two planck’s

1.918 × 10

6.626 ×10 1.918 × 10

And

−34

from

that

J

whereas the accepted planck’s constant value is makes

a

difference

of

around

constant value, there are not much difference which shows a bit of accuracy while measuring the trials of backing voltage. Thus, the percentage error of planck’s constant is approximately 0%. Part C: Relationship between Illumination and Current The data table given below records the dimension of apertures of different size which will also be used to determine the area of each aperture size and also the phototube current. Note that the area of the aperture is determined using the given diameter. Diameter (m)

Area ( ×10−3 m 2 )

Phototube Current (µA)

0.007

0.154

0.29

0.010

0.314

0.40

0.014

0.616

0.48

0.020

1.257

0.54

And then using the above table, the graph of the current through the tube against the aperture area is sketched as shown below.

Graph of Tube Current Vs Aperture Area 0.6 0.5

Phototube Current (µA)

0.4 0.3 0.2 0.1

-0.1

0

0.1

0.3

0.5

0.7

0.9

1.1

1.3

1.5

-0.1

Aperture Area (×10^(−3) � ^2)

The graph above shows a smooth curve of the relationship between the phototube current and the aperture area. It clearly shows by observation of the graph drawn above that the current is directly proportion to the aperture area so when the size of aperture increases, the current passing through does increase as well. Hence, the relationship can be expressed or denoted as Current is directly proportional ¿ Aperture → I ∝ A

.

Discussion Questions: The photoelectric threshold frequency of a metal is the frequency of the incident light which arises during the emission of the electrons from the surface of that particular metal. The required energy to initiate the photoelectric effect from the surface of the metal is the work function and it is directly proportional to the threshold frequency. Threshold frequency is the ratio of work function and planck’s constant which is expressed as when work function increases, the threshold frequency will also increase.

f o=

ϕ . Hence, h

All characteristics of light described by quantum theory and according to the experimental analysis above, the data shows the release of light energy from the energy source packs of the frequency specifically due to the emission of electrons from the metal surface. And in comparison, to Einstein’s theory which states that light travels in a group of energy where each energy group called photon and all individual photons has the same amount of energy quantity as the product of frequency and planck’s constant. The graph of backing voltage against the frequency showcases a clear image of quantum nature theory of light. The backing voltage depends on the light frequency due to the relationship expressed in the formula of electric-voltage eV =hf −ϕ where hf is the energy source and work function ɸ is minimum energy of metal. A clear image displaying the relationship is shown in the graph of backing voltage against frequency above. However, there is no relationship between the backing voltage and intensity but the frequency of light does have a relationship with intensity. Hence, light frequency is directly proportional to both backing volts and intensity. The photoelectric effect can be observed in other metal surfaces with the equivalent incident light condition depending on the type of the metal. However, it is not necessary because all metals have its’ own work function in which the minimum energy needed to free electrons from its surface. For instance, metals with a low work function such as potassium can display photoelectric the moment it is illuminated by lower energy photons. On the other hand, metals with higher work function needs more active photons.

Conclusion: To conclude, the experiment was successfully completed where all the required tables, graphs and calculations were all displayed above. The photoelectric effect was able to be observed from the first part of the experiment due to the backing voltage of 0V and tube current of 0.47µA. In the second part, the planck’s constant of 1.918 × 10−28 J

was able to

determine from the graph of stopping or backing voltage versus frequency and in comparison, with the accepted value of the backing voltage of

6.626 ×10

−34

J , the

percentage error is approximately 0%. And in the third part, the relationship of the current and illumination was able to be observed from the graph of phototube current versus aperture area. Thus, there are certain experimental errors occur during the process of the

experiment, these errors happen due to old electrical components and the lack of accuracy in data readings.

References: Kumar, D. S. et al., 2009. PH103: Quantum & Electrical Physics, Suva: Physics Department....