Transcript Blackbody Radiation

Basic Properties of Stars - 4 §3.4-3.6
Colours of stars and blackbody radiation
Colours of stars
Stars have colours. Why?
Colours of stars
Stars have colours. Why?
Its not due to their redshift!!
Blackbody Radiation
Betelgeuse: red
Surface Temp = 3600K
Rigel: blue-white.
Surface Temp = 13,000K
Orion
Blackbody Radiation History
In 1792, Thomas Wedgewood first observed that all his
ovens glowed “red-hot” at the same T, regardless of size,
shape or materials.
All objects T > 0 K emit radiation.
Below 800 K in the IR, 800 to 1000 K detected in optical
By 3000 K white hot - as T goes up
Spectrum shifts to shorter wavelengths & power increases
Perfect emitter absorbs all light it receives and reradiates it
- called a blackbody.
Blackbody Radiation History
Perfect blackbody is an idealization, but closely
approximated as below.
Cavity constant Temp 
J. Stefan 1879 found
Small relation between total
hole power emitted and T
P=T4
=5.67 x 10-8 W m-2 K-4
Blackbody Radiation
Blackbody spectrum depends only on T of source.
Blackbody Radiation
Blackbody spectrum depends only on T of source.
As T increases, max decreases and B()d increases
maxT = d (Wien’s Law) and d = 2.898 x 10-3 m K.
Blackbody Radiation and Quanta
By 1900, Max Planck found empirical formula for
blackbody curve.
B(T) = (a-5) / e b/T -1 and he tried to derive the
constants a and b.
Problem is shown on following slides
Standing waves
Standing waves
Blackbody radiation
To circumvent this problem, Planck assumed that a standing E-M
wave could not acquire any arbitrary amount of energy, but only
allowed values that were multiples of a minimum wave energy.
This quantum is given by h (or hc/), where h is constant = 6.63 x
10-34 J sec (Planck’s constant).
Higher  (shorter ) of wave, greater minimum energy.
Short , high  waves cannot contain even 1 quantum!
The Planck Function
B(T) = energy emitted per second, per unit wavelength
interval d at wavelength , per unit area into a unit
solid angle by a blackbody of temperature T (whew!)
B(T) = (2hc2/5)(1/ehc/kt - 1) w m-2 m-1 sterad-1
Where c = speed light = 3 x 108 m s-1
k = Boltzmann constant = 1.38 x 10-23 J K-1
h = Planck’s constant = 6.63 x 10-34 J s
Only variable in the Planck function is T
In terms of frequency B(T) = (2h3/c2)(1/eh/kt - 1)
(require Bd =-Bd - as  decreases with increasing ,
d/d = -c/2, B =-B d/d =B c/2)
The Planck Function
How does Planck function behave in the limits of very
high and very low frequency? i.e h/kt >> 1 and << 1
Set h/kT = x  B(T) = (2h3/c2)/(ex - 1)
For x >> 1, ex -1 = ex so B(T) = (2h3/c2) e-h/kT
This called Wien approximation
For x << 1, ex = 1 + x + x2/2 + x3/6 + ……… xn/n! = 1 + x
Thus B(T) = (2h3/c2)(1/x) = 2kT2/c2
Thus log B(T) = 2 log  + log T + constant
Blackbody radiation
BB radiation - total intensity
B(T) = (2h3/c2)(1/eh/kt - 1)
The total intensity emitted by the BB is the integral B(T) = B(T)d
= B(T)d = (2h3/c2)(1/eh/kt - 1)d where the integral goes from
0 to .
Substitute x = h/kT, so that d = (kT/h)dx.
Then B(T) = (2hk4T4/c2h4)(x3/ex -1)dx
Integral just a real number so that B(T) = AT4 with
A = 2k44/15c2h3
So B(T)  F  T4 (F = T4). This is the Stefan-Boltzmann Law
 = 5.67 x 10-8 w m-2 K-4
Blackbody Radiation
Relation between max and T is known as Wien’s Law
maxT= 0.002897755 m K = 0.290 cm K.
For a spherical source:
F = L / 4R2, (R radius circle
surrounding source)
from S-B Law
F = T4 so L = 4R2Te4
Te is the effective T as stars
are not perfect BB radiators T of BB that puts out same
energy as the star
Blackbody Radiation
A simple problem:
Lsun = 3.839 x 1026 W and its radius is Rsun =
6.955 x 108 m.
(a) What is Te of Sun?
(b) Where does Sun’s flux peak?
(c) Any relation to the sensitivity of human eye?