Transcript Problems Part 2

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?
λMax = (0.29 cm / T)
T in K
λMax = (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.
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?
λApparent / λTrue = 1 + (speed of source / wave speed)
λApparent / λTrue = 1 + (.5 c / c) = 1.5
λApparent = 1.5 (600 nm) = 900 nm
No, infrared
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?
λApparent / λTrue = 1 + (speed of source / wave speed)
λApparent / λTrue = 1 + (.5 c / c) = 1.5
λApparent = 1.5 (600 nm) = 900 nm
No, infrared
Examples
When the first astronauts landed on the moon, how long did it take radio
transmissions to reach the earth?
d=vt=ct
t = d / c = (3.84 x 108 ) / (3.00 x 108 m/sec) = 1.28 sec
Examples
If the sun just blew up, how long would it be before we would see the
explosion?
d=vt=ct
t = d / c = (1.5 x 1011 m) / (3.00 x 108 m/sec) = 500 sec = 8.33 minutes
Examples
Line
 (nm)
Line
 (nm)
Ly
122
UV
H
656
Visible
Ly
103
UV
H
486
Visible
Ly
97
UV
H
434
Visible
Examples
Show that the wavelength of the Hβ line is 486 nm.
EHβ = 13.6 (1/22 – 1/42) 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 1014 sec-1
f = 6.18 x 1014 sec-1 = c / λ
λ = (3 x 108 m/sec) / (6.18 x 1014 sec-1) = 485 x 10-9 m = 485 nm
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 3rd excited state?
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 3rd excited state?
Energy
-13.6 /42
-13.6 /32
-13.6 /22
-13.6 /12
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 3rd excited state?
Energy
-13.6 /42
-13.6 /32
-13.6 /22
-13.6 /12
Answer – SIX ! The red transition is redundant