Transcript Tutorial: Continuous Spectra

Spectral Types – Continuous Spectra
Stellar Spectra
Since for a photon
E=hf
The energy carried by the photon determines the position of the
electromagnetic wave on in the electromagnetic spectrum.
If the resulting frequency is within the visible window, the energy of the
photon will determine the color of the electromagnetic wave.
Wein’s Law
An star’s surface temperature is 5000 K. The star is moving toward the Earth
at a speed of 3 x 107 m/sec. What is the peak wavelength for the continuous
spectrum of this star at the surface of the star? What color is the star to a
observer in a space capsule who stationary with respect to the star?
According to Wein’s Law
Wavelength of Peak Emission =
0.29
Temperature in K
Therefore,
λPeak = 0.29/50000 cm = 5.8 x 10-7 cm = 580 nm.
According to our discussions in class, this star would be yellow.
cm
Wein’s Law
An star’s surface temperature is 5000 K. The star is moving away from the
Earth at a speed of 3 x 107 m/sec. What is the peak wavelength for the
continuous spectrum of this star as measured from the earth? What color is
the star to a observer on the Earth?
From the previous example, the peak wavelength for the continuous spectrum at
the surface of the star is
λPeak = 0.29/50000 cm = 5.8 x 10-7 cm = 580 nm.
However, since the star is moving away from the Earth, it’s wavelengths will be
Doppler shifted. Therefore, measurements from the Earth yield a peak wavelength of
λApparent / λTrue = λApparent / 580 nm = 1 +(3 x 107 m/sec / 3 x 108 m/sec)
→ λApparent = 1.1 (580 nm) = 638 nm.
According to our discussions in class, this star would be red when observed from the
Earth.
Wein’s Law
If the star in the previous example was moving away from the earth at a speed
of 6 x 107 m/sec, what is the peak wavelength for the continuous spectrum of
this star as measured from the earth? What color is the star to a observer on
the Earth?
From the previous example, the peak wavelength for the continuous spectrum at
the surface of the star is
λPeak = 0.29/50000 cm = 5.8 x 10-7 cm = 580 nm.
However, since the star is moving away from the Earth, it’s wavelengths will be
Doppler shifted. Therefore, measurements from the Earth yield a peak wavelength of
λApparent / λTrue = λApparent / 580 nm = 1 +(6 x 107 m/sec / 3 x 108 m/sec)
→ λApparent = 1.5 (580 nm) = 870 nm.
According to our discussions in class, this star would not be visible to optical
telescopes. It would, however, be visible to a telescope collecting information in the
infrared portion of the electromagnetic spectrum.
Stefan’s Law
What is the power emitted per unit area of the star in the previous example?
According to Stefan’s Law,
Power Emitted per unit Area =  T4
 = 5.67 x 10-8 W / m2 – K4
(Stefan-Boltzmann constant)
Note: A Watt (W) is a Joule/Sec
Therefore,
Power/Area = (5.67 x 10-8 W/m2 – K4 ) (5000 K)4 = 3544 x 104 W/m2 = 3.544 x 107 W/m2 .
Stefan’s Law
What is the total power emitted over the entire surface area of the star in the
previous example? Assume the radius of the star is the same as that of the
sun, 7 x 105 km.
From the previous example,
Power/Area = (5.67 x 10-8 W/m2 – K4 ) (5000 K)4 = 3544 x 104 W/m2 = 3.544 x 107
W/m2 .
The surface area of a star is given by 4πR2, and therefore for this star
4πR2 = 4 (3.14) (7 x 108 m)2 = 6.154 x 1018 m2 .
As a result, the total power emitted by the star is
P = (Power/Area) (Area) = (3.544 x 107 W/m2 ) (6.154 x 1018 m2 ) = 2.18 x 1026 W.