The sun can be considered an ideal blackbody radiator at a temperature of 5900K.

Question

The sun can be considered an ideal blackbody radiator at a temperature of 5900K. (a) What is the peak wavelength of the distribution? (b) What is the energy in Joules of a peak blackbody photon at this wavelength? (c) Is this photon in the visible band of the electromagnetic spectrum (show why)? (d) A newly discovered star in the Bug Nebula is the hottest star in the galaxy, 35 times hotter than the temperature of the sun.(e) What is the peak wavelength, and frequency of this blackbody distribution?

Explanation / Answer

ANSWER

THIS WILL HELP YOU

A certain source emits radiation of wavelength 500.0 nm. What is the energy, in kJ, of one mole of photons of this radiation?

Solution:

1) Convert nm to m:

2) Determine the frequency:

(5.000 x 10 ¯7 m) (x) = 3.00 x 108 m/s

x = 6.00 x 1014 s ¯1

3) Determine the energy:

x = (6.626 x 10 ¯34 J s) (6.00 x 1014 s ¯1)

x = 3.9756 x 10 ¯19 J

Important point: this is the energy for one photon.

4) Determine energy for one mole of photons:

239.4 kJ/mol

Comment: if you wished to do a direct calculation, you could use this equation:

Just make sure that the units for c and λ are the same.

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