Uploaded by Nur Arinah Mohamad

LAB REPORT EXPERIMENT 1 (Blackbody Radiation)

PHY310/March 2020 – July 2020
PHY 310
Experiment No.
Program / Group
: AS1204G
Date of Experiment
: 20/04/2020
: Dr. Mohd Nazari Abu Bakar
Matrix No
Nur Arinah Binti Mohamad
1. To establish relationship between temperature, Wien’s Law and
2. To create understanding of how thermal spectra can be used to evaluate the
temperature of an object.
Blackbody radiation is an important phenomenon when studying thermal radiation
and electromagnetic radiation energy transfer in all wavelength bands. Blackbody
itself is an ideal object that absorb all incident light hence the name ‘black’. This ideal
radiation absorber is used as a standard to compare radiation emitted by real physical
bodies. Using known formalities such as Stefan’s Law and Wien’s Displacement Law,
the energy transferred to environment and the peak wavelength can be calculated.
The peak emission of this radiation depends on the temperature. The hotter an object
is the more radiation it emits at shorter wavelengths with more energy i.e. at higher
1. The link https://phet.colorado.edu/sims/html/blackbody-spectrum/latest/blackbody
spectrum_en.html is clicked to open the simulation.
2. Once the simulation is open, try to play with it.
3. The temperature is set to the temperature of the Sun. At this temperature, the peak
wavelength is found and the colour (in the visible range) is determined. Compare this
colour and the colour of the sunlight. Is there is any difference? Why?
4. The camera button is clicked to save the Sun spectrum. Next, the temperature
slider is changed to 7000 K and 8000 K. The temperature, peak wavelength, and
power intensity at each temperature is recorded. These values is put in the table by
using the Wien’s Law and Stefan-Boltzmann’s Law equations. Then, the error
percentage is calculated.
Peak wavelenght,  peak (nm)
Temperature, T (K)
Intensity, I (W/m4)
6.42 107
6.42 107
1.36  108
1.36  108
2.32 108
2.32 108
Peak wavelenght, λpeak
Using Wien’s Law
λmaxT = 2.898 x 10-3m.K
1. T= 5800K
λmax(5800K) = 2.898 x 10-3m.K
λmax= 2.898 x 10-3m.K / 5800K
= 5.0 x 10-7m / 10-9
λmax= 500 nm
2. T= 7000K
λmax(7000K) = 2.898 x 10-3m.K
λmax= 2.898 x 10-3m.K / 7000K
= 4.14 x 10-7m / 10-9
λmax= 414 nm
3. T= 8000K
λmax(8000K) = 2.898 x 10-3m.K
λmax= 2.898 x 10-3m.K / 8000K
= 3.62 x 10-7m / 10-9
λmax= 362 nm
Intensity, I (W/m4)
Using Stefan-Boltzman Law
S= σT4
1. T= 5800K
S= 5.6703 x 10-8 W/m2 k4 (5800K)4
= 6.42 x 107 W/m2
2. T= 7000K
S= 5.6703 x 10-8 W/m2 k4 (7000K)4
= 1.36 x 108 W/m2
3. T= 8000K
S= 5.6703 x 10-8 W/m2 k4 (8000K)4
= 2.32 x 107 W/m2
Analysis Data
Percentage Error
%error 
| simulated  calculated |
 100
From graph λ simulation vs λ calculated,
where, simluated  calculated
| simulated  calculated |
 100
%error  0%
%error 
From graph I simulation vs I calculated,
where, Isimluated  Icalculated
| simulated  calculated |
 100
%error  0%
%error 
1. Compare the colour that you found in method no.3 and the colour of the sunlight. Is
there is any difference? Why?
The sun's spectrum peaks in the green wavelengths.Therefore, the colour shows it is
green while the color of sunlight that we see is yellow-red. There is a huge different.
This is because,the truth to be told that when we're above the sensible atmosphere
we see sunlight as white because the sun emits all colors of the rainbow more or less
evenly and in physics, we call this combination "white". But our eyes don not isolate
the peak color so we could not see the green color appears by our naked eyes. If
sunlight were purely green, then everything outside would look green or dark. The sun
emits all colors of visible light, and in fact produces all frequencies of electromagnetic
waves except gamma rays. This includes radio waves, microwaves, infrared waves,
visible light, ultraviolet waves, and X-rays. The sun emits all these colors because it is
a thermal body and emits light through the process of thermal radiation. The
atmosphere scatters short wavelengths much more effectively than long wavelengths.
This is why we get those lovely red sunsets where all the air between our eyes and
the setting Sun scatters away so much of the blue end of the "rainbow" that what's left
is largely red.
2. If the body of the human body is around 37°C, can we see the light coming from the
body? Why? Use the simulation to support your answer.
We cannot see the light coming from our body as if the human body temperature is
310.15K. This can explained through the simulation that have been conducted. Based
on the diagram of the simulation, its peak wavelenght lies in the infrared range.
Human vision can see red light wavelengths as long as about 700 nm. The infrared
wavelengths are generally defined as being between 700 and 100,000 nm. Infrared
light, is electromagnetic radiation with wavelengths longer than those of visible light. It
is therefore generally invisible to the human eye, although infrared at wavelengths up
to 1050 nm from specially pulsed lasers can be seen by humans under certain
conditions only.
In conclusion, from this experiment we can see that when the temperature increases,
its peak wavelenght and intensity decreases. I have better understanding on
blackbody radiation while doing this simulation as I can understand clearer that
different temperature has its different peak wavelenght where the spectrum reaches
its highest intensity.