Please I need this by 1pm I have attached the picture for question 1
1. It just so happens that regardless of the material, when objects are heated up they will start to glow and change colors at near identical temperatures. The plot that you see is called a blackbody spectrum. This plot tells us the intensity or the “amount” of light that an object will emit at different wavelengths (or “colors”). The visible wavelengths are marked by their colors on the plot. To the right of the visible band is lower energy infrared light. To the left of this band is higher energy ultraviolet (UV) light.
Click the + button that is to the left of the intensity scale (far left side of the screen) such that the top of the scale is at .001. (in the picture above the top of the scale says 100).
Now use the temperature slider to the right, and take the temperature all the way down to 300 Kelvin (80 Fahrenheit).
Now slowly begin to raise the temperature. At approximately what temperature would a heated material (metal, wood, etc.) begin to give off visible light at a deep red color?
Note: This will be the temperature where your spectrum first begins to come off of the wavelength axis in the visible region, and so is giving off a small amount of red light.
- 500 K (440 Fahrenheit)
- 1050 K (1430 Fahrenheit)
- 1800 K (2780 Fahrenheit)
- 2500 K (4040 Fahrenheit)
2. Click the – button that is to the left of the intensity scale to zoom out such that the top of the scale is at 10.
Move the temperature slider to that of a light bulb. The red part of the thermometer on the far right should just be touching the line marked light bulb. At approximately what temperature does the filament in a household light bulb operate?
Note: This is written in blue in the simulation.
- 660 K (728 F)
- 1800 K (2780 F)
- 3000 K (4940 F)
- 5700 K (9800 F)
3. What type of light does this light bulb produce most (i.e. at what wavelength does the spectrum have maximum intensity)?
- Infrared light
- Red visible light
- Violet visible light
- Ultraviolet light
4. Click the – button that is to the left of the intensity scale to zoom out such that the top of the scale is at 100.
Move the temperature slider to that of the Sun. The red part of the thermometer on the far right should just be touching the line marked Sun. Approximately what temperature is the surface of the Sun?
- 2100 K (3320 F)
- 4500 K (7640 F)
- 5700 K (9800 F)
- 9800 K (17,180 F
5. Based on the simulation, what type of light does the Sun produce the most?
- Infrared light
- Green visible light
- Orange visible light
- Ultraviolet light
6. Relative to the peak intensity in the Sun’s spectrum, the Sun emits nearly equal amounts of light across the entire visible part of the EM-spectrum. This is demonstrated by the star shaped symbol at the top of the simulation being white. Therefore, if you look at the Sun when it is directly overhead on a clear day, it will appear white.
Click the – button that is to the left of the intensity scale to zoom out such that the top of the scale is at 316.
Use the star shaped symbol above your graph and to the right of the blue, green, and red dots to estimate the temperature at which something will begin to glow blue. At approximately what temperature does the object gain a faint blue tint?
Note: This will also be the temperature where the max intensity of the objects spectrum is in the blue portion of the visible spectrum.
- 3000 K (4940 F)
- 6600 K (11,420 F)
- 7900 K (13,760 F)
- Object cannot glow blue at any temperature.
7. Note that in the above question, although the object still emits all colors of visible light, it appears blue now instead of white because of the significant difference in the intensity or amount of blue light radiated versus the amount of red light emitted.
Click the + button that is to the left of the intensity scale to zoom in such that the top of the scale is at 1. Now slowly decrease the temperature from 5000K down to 300K (room temperature).
Notice how the entire spectrum decreases in intensity and moves to the right into the infrared region. Even though the spectrum appears completely flat, objects at room temperature and below also emit their own light. If our eyes could detect infrared light, we would be able to see in the dark with warmer objects being brighter than others.
In the introduction of this activity, we mentioned the temperature of your home on hot and cold days. Your body is kept warm in your home primarily by two ways: by direct contact with the air around you and by absorbing infrared light that is radiated from the walls. As you have seen in this activity, the light that is radiated from an object depends almost solely on the temperature of the object. Based on what you have learned here, what is one reason for feeling warmer in your house on a summer day versus a winter day even though your thermostat is set the same?
- The walls of the house are warmer during the summer. Therefore, they radiate more infrared light that can serve to warm our body.
- The walls of the house are warmer during the summer. Therefore, they radiate more visible light that can serve to warm our body.
- The walls of the house are warmer during the summer. Therefore, they radiate more ultraviolet (UV) light that can serve to warm our body.
- The temperature of the walls of the house has no effect on the light they radiate.
8. Since we cannot physically collect data from stars and most other objects in the universe, almost all of the information we obtain from the universe comes from analyzing the light, or spectra, from those objects.The study of light is known as spectroscopy.
As we have seen in this simulation, every blackbody emits light with an easily identified pattern known as the blackbody curve. This is the particular way the total light emitted by a blackbody varies with its frequency. The exact form of the curve depends only on the body’s temperature. Since we can treat stars as blackbodies, this is incredibly useful in astronomy that shows us that the color of a star is also indicative of its temperature.
Use the simulation to determine the surface temperature of the following star:
Betelgeuse is a red supergiant star in the constellation Orion.
Knowing that Betelgeuse has peak intensity in the red and infrared wavelengths, adjust the intensity scale and temperature until you can determine the approximate surface temperature of the star.
- 3500 K
- 4800 K
- 7700 K
- 11,000 K
9. In this equation:
λ(max)= peak wavelength (cm)
T = temperature (K)
Based on what you have seen in the simulation and your knowledge of proportionality relationships learned this month, what is the relationship between temperature and peak wavelength?
- They are directly proportional.
- They are inversely proportional.
- They are exponentially proportional.
- They are unrelated
10. Use Wien’s Law to calculate the peak wavelength of Betelgeuse, based on the temperature found in Question #8.
Note: 1 nanometer (nm) = .0000001 centimeters (cm)
- 208 nm
- 400 nm
- 828 nm
- 1800 nm