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Planck distribution

In the previous section we discussed light and matter at equilibrium in a two-level quantum system. For the remainder of this section we will be interested in light and matter which are not at equilibrium. In particular, laser light is completely different from the thennal radiation described at the end of the previous section. In the first place, only one, or a small number of states of the field are occupied, in contrast with the Planck distribution of occupation numbers in thennal radiation. Second, the field state can have a precise phase-, in thennal radiation this phase is assumed to be random. If multiple field states are occupied in a laser they can have a precise phase relationship, something which is achieved in lasers by a teclmique called mode-locking Multiple frequencies with a precise phase relation give rise to laser pulses in time. Nanosecond experiments... [Pg.225]

This is the Planck distribution function. The themial average energy in theyth mode is (including the zero point energy)... [Pg.409]

Figure 4.24. The Planck distribution law spectral radiance of blackbody radiation as a function of temperature and wavelength. (After Touloukian and DeWitt (1972). Plenum Press.)... Figure 4.24. The Planck distribution law spectral radiance of blackbody radiation as a function of temperature and wavelength. (After Touloukian and DeWitt (1972). Plenum Press.)...
The importance of spontaneous emission is that we can now understand the evolution of the system matter + radiation in thermodynamic terms as the evolution toward a state of maximum entropy which is realized when the matter reaches the Boltzmann distribution and the photons the Planck distribution. [Pg.13]

Spectral distribution of blackbody radiation. The family of curves is called the Planck distribution after Max Planck, who derived the law governing blackbody radiation. Note that both axes are logarithmic. [Pg.427]

The exitance (power per unit area per unit wavelength) from a blackbody (Box 20-1) is given by the Planck distribution ... [Pg.451]

Planck distribution Equation giving the spectral distribution of blackbody radiation ... [Pg.698]

Exercise. Replace in (5.6) the factor 1// by the Planck distribution and find in this way the quantum mechanical equivalent of (5.5). [Pg.67]

LI The Planck Distribution of Black-body Radiation. The Planck relationship between the energy of the photon and the frequency of monochromatic light leads to the equation of distribution of the intensity of light as a function of frequency (or wavelength)... [Pg.24]

Thus u = 0 in thermal equilibrium (s=0), when (4) becomes a Planck distribution, and... [Pg.215]

The effect of the gray approximation on emissivity and emissive power of a real surface is illustrated in Fig. 12 27. Note that the radiation emission from a real surface, in general, differs from the Planck distribution, and the emission curve may have several peaks and valleys. A gray surface should emit as much radiation as the real surface it represents at the. same temperature. Therefore, the areas under the emission curves of the real and gray surfaces must be equal. [Pg.697]

This implies that the higher-energy modes are less populated than what is implied by the equipartion principle. Substituting this value, rather than kT, into the Rayleigh-Jeans formula (1.21), we obtain the Planck distribution law... [Pg.178]

From the Planck distribution law one can calculate the wavelength at which p X) is a maximum at a given T. The result agrees with the Wien displacement law with... [Pg.178]

Before proceeding to explore the implications of the Planck distribution, we need to check whether it reduces to the classical expression under the appropriate conditions. It is always important to check whether new concepts can be... [Pg.122]

The distribution function (404) is much greater than the Planck distribution function... [Pg.456]

Q.7.4 Show that (a) the Rayleigh-Jeans law is a special case of Planck distribution law for the blackbody spectrum. Show also that (b) the Wein displacement law can be derived from Planck s distribution law. [Pg.35]

An interesting problem is the field evolution in a cavity that was initially in the equilibrium state at a finite temperature, when the initial occupation numbers were given by the Planck distribution v = [exp(pw) — 1] 1. Let us consider two limit cases. The first one corresponds to the low-temperature approximation v = exp (—(] ). Then the occupation number of the mth mode is merely the coefficient at vm in the expansion (61) with u = exp( (5). Using the well-known generating function of the Legendre polynomials Pm(z) [Ref. 269, Eq. 10.10(39)], one can obtain the following expression (for y = 0) ... [Pg.331]

This depends only on the invariant minimal and maximal variances um and vm. Note that the argument of the polynomial in (151) is always outside the traditional interval (—1,1) (in particular, this argument is purely imaginary if 2um < 1), being exactly equal to 1 for the nonprincipal modes with um = vm = Jrm+j, when formula (151) transforms to the time-dependent Planck distribution... [Pg.353]

Equation 2.6 Planck distribution of a black body radiator... [Pg.31]

Figure 7.3a represents the Planck distribution for blackbody spectral emissive power with E-i p / as a function of XT. The band fraction of emitted energy in the region from 0 to XT is equal to the shaded area, which is expressed as and shown in Figure 7.3b. About a quarter of the emitted energy is at wavelengths shorter than nd nearly 95% of the emitted energy is distributed between and The spectral distribution of solar radiation can be... Figure 7.3a represents the Planck distribution for blackbody spectral emissive power with E-i p / as a function of XT. The band fraction of emitted energy in the region from 0 to XT is equal to the shaded area, which is expressed as and shown in Figure 7.3b. About a quarter of the emitted energy is at wavelengths shorter than nd nearly 95% of the emitted energy is distributed between and The spectral distribution of solar radiation can be...
The total emissivity can be calculated from the spectral emissivity using the Planck distribution see Equation (7.9). Using Equation (7.7), it can be shown that at 300 K nearly 98% of the blackbody radiation is at wavelengths longer than 6 jm therefore, the total emissivity at 300 K is equal to the spectral emissivity beyond 6 pm (see Figure 7.8). Hence, e 3 = 0.3. Because the surface is diffuse, from Kirchhoff s law, The total absorptivity can be calculated from Equation (7.11). The... [Pg.575]

In a combustion experiment of a luminous hydrocarbon flame, it is desired to measure the total irradiation received by a sensing device. The bright yellow color of the flame indicates that the radiation emitted by the small soot particles obeys the Planck distribution. The optical sensing device is a photomultiplier tube (PMT) which converts the photon interaction at the inlet (photocathode) to an electric current at the outlet (anode). The PMT has 3 cm2 of active sensor area facing the flame and is placed 2 m away from the flame axis. [Pg.420]

P8.12 (a) With a little manipulation, a small-wavelength approximation of the Planck distribution can be... [Pg.155]


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