1)The temperature of the visible surface of the sun, the photosphere, is around

Question

1)The temperature of the visible surface of the sun, the photosphere, is around 5610 C. (The temperature of the inside of the sun is much hotter.). At what frequency does the Sun emit its peak radiation?

fpeak = ? /sec

2)What is the wavelength of this radiation?

peak = ?nm

3)The human eye has evolved so it can detect electro-magnetic radiation with wavelengths of from roughly 380m to 720m. This is the visible light spectrum. This would appear to make sense based on the location of the peak of the radiation spectrum from the sun. The tungstan filement of standard incandescent household light bulb operates at a temperature of around 2620 C. At what frequency does the bulb emit its peak radiation?

fpeak =? /sec

4)What is the wavelength of this radiation?

peak =? nm

5)This isn't very good, right? The peak is well outside the visible spectrum so a lot of the radiation produced by the bulb is wasted. (Unless of course, you're using the light bulb to heat up something.) We can fix this by making the bulb more like the sun. By surrounding the filament of the bulb with halogen gas (or gasses from the same chemical group like iodine or bromine), we can allow the filament to operate at a higher temperature. The filament in halogen light bulb operates at a temperature of around 3050 C. At what frequency does this bulb emit its peak radiation?

fpeak =? /sec

6)What is the wavelength of this radiation?

peak =? nm

Explanation / Answer

(1)

The wavelength of peak radiation is given by,

T = b, where b is Wien's constant with b = 2.898 x 10 -3m-K

or = b/T = 2.898 x 10 -3m-K/(5610+273)K

or, = 492.6 nm

So frequency is f = c/ = (3X108m/s)/(489.5X10-9m)

or f = 6.09X1014s-1

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(2)

As we saw above the wavelength is,

= 492.6 nm

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(3)

for tungsten filament,

= b/T =2.898 x 10 -3m-K/(2620+273)K

or, = 1001.72 nm

So the frequency is

f =  c/ = (3X108m/s)/(995.5X10-9m)

or f = 2.99X1014 s-1

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(4)

As we saw above the wavelength is,

= 1001.72 nm

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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