Fx

x2ex

This vanishes, when 3 — 3e—x — x = 0. The solution of this equation is x = 2.821. Hence cT hc

By definition, x is always strictly positive. Hence f '(x) can be zero only if the factor 5 — 5e—x — x is zero. This equation cannot be solved analytically. Instead we write the equation as x = 5 — 5e—x and solve it by iteration:

x0 = 5 , (this is just a guess) xx = 5 — 5e—x0 = 4.96631,

Thus the result is x = 4.965. The Wien displacement law is then hc

In terms of frequency Planck's law is

Note that the wavelength corresponding to vmax is different from Amax. The reason is that we have used two different forms of Planck's function, one giving the intensity per unit wavelength, the other per unit frequency.

Example 5.4 a) Find the fraction of radiation that a blackbody emits in the range [Ai,A2], where Ai and A2 ^ Amax. b) How much energy does a 100 W incandescent light bulb radiate in the radio wavelengths, A > 1 cm? Assume the temperature is 2500 K.

Since the wavelengths are much longer than Amax we can use the Rayleigh-Jeans approximation BA(T) « 2ckT/A4. Then

A1 A1

J A4

and hence B'

Now the temperature is T = 2500 K and the wavelength range [0.01 m, i), and so

25003 0.0i3

x v max

It is quite difficult to listen to the radio emission of a light bulb with an ordinary radio receiver.

By substituting different values for Tc, we find that Tc = 7545 K satisfies our equation.

Example 5.5 Determination of Effective Temperature

The observed flux density of Arcturus is

Interferometric measurements give an angular diameter of a = 0.020". Thus, a/2 = 4.85 x 10-8 radians. From (5.26) we get

=4300K.

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