1. Examples A star radiates as a blackbody at a temperature of 1700 K. At what wavelength does the peak of the blackbody spectrum occur? If you were to look at the source, what color would it appear to be? λ Max = (0.29 cm / T)T in K λ Max = (0.29 cm / 1700) = 1.7 x 10 -4 cm = 1.7 x 10 -6 m = 1700 nm This is outside of the visible spectrum in the IR region (long wavelength). Therefore, one would expect that only the long wavelength end of the visible spectrum would be present. The star would be red.

2. Examples A particular object is moving away from the earth at 50% of the speed of light. What is the wavelength as measured on the earth of an electromagnetic wave which leaves the source with a wavelength of 600 nm? Can this wave be observed with an optical telescope? If not, in what part of the electromagnetic spectrum will it be observed? λ Apparent / λ True = 1 + (speed of source / wave speed) λ Apparent / λ True = 1 + (.5 c / c) = 1.5 λ Apparent = 1.5 (600 nm) = 900 nm No, infrared

3. Examples A particular object is moving away from the earth at 50% of the speed of light. What is the wavelength as measured on the earth of an electromagnetic wave which leaves the source with a wavelength of 600 nm? Can this wave be observed with an optical telescope? If not, in what part of the electromagnetic spectrum will it be observed? λ Apparent / λ True = 1 + (speed of source / wave speed) λ Apparent / λ True = 1 + (.5 c / c) = 1.5 λ Apparent = 1.5 (600 nm) = 900 nm No, infrared

4. Examples When the first astronauts landed on the moon, how long did it take radio transmissions to reach the earth? d = v t = c t t = d / c = (3.84 x 10 8 ) / (3.00 x 10 8 m/sec) = 1.28 sec

5. Examples If the sun just blew up, how long would it be before we would see the explosion? d = v t = c t t = d / c = (1.5 x 10 11 m) / (3.00 x 10 8 m/sec) = 500 sec = 8.33 minutes

6. Examples Line  (nm) Line Ly  122UV HHHH656Visible Ly  103UV HHHH486Visible Ly  97UV HHHH434Visible

7. Examples Show that the wavelength of the Hβ line is 486 nm. E Hβ = 13.6 (1/2 2 – 1/4 2 ) eV = 13.6 (3/16) eV = 2.55 eV. 2.55 eV (1.6 x 10 -19 J/eV) = 4.08 x 10 -19 J = hf (4.08 x 10 -19 J ) / (6.6 x 10 -34 J – sec) = 6.18 x 10 14 sec -1 f = 6.18 x 10 14 sec -1 = c / λ λ = (3 x 10 8 m/sec) / (6.18 x 10 14 sec -1 ) = 485 x 10 -9 m = 485 nm

8. Examples How many different energy photons can be emitted by a gas of hydrogen atoms if (by magic) all of the electrons start in the 3 rd excited state?

9. Examples How many different energy photons can be emitted by a gas of hydrogen atoms if (by magic) all of the electrons start in the 3 rd excited state? Energy -13.6 /1 2 -13.6 /2 2 -13.6 /3 2 -13.6 /4 2

10. Examples How many different energy photons can be emitted by a gas of hydrogen atoms if (by magic) all of the electrons start in the 3 rd excited state? Energy -13.6 /1 2 -13.6 /2 2 -13.6 /3 2 -13.6 /4 2 Answer – SIX ! The red transition is redundant