Looking for a Tutor Near You?

Post Learning Requirement » x
Ask a Question
x

Choose Country Code

x

Direction

x

Ask a Question

x

Hire a Tutor
Delhi

Black Body Radiation

Home > PowerPoint Slides > Black Body Radiation
Published in: Physics
29,803 Views

Black Body Radiation-Modern Physics

Akhilesh K / Lucknow

4 years of teaching experience

Qualification: M.Sc (NIT Rourkela - 2019)

Teaches: All Subjects, English, Mathematics, Science, Chemistry, Physics, Algebra, IIT JEE Mains, AIPMT, NEET

Contact this Tutor
  1. Modern physics ' Black Body ' Black Body Radiation Spectra ' Various Theories to Explain Black Body Radiation Spectra
  2. Blackbody Radiation When matter is heated, it emits radiation. A blackbody is a cavity in a material that only emits thermal radiation. Incoming radiation is absorbed in the cavity. e 20m - • Blackbody radiation is theoretically interesting because the radiation properties of the blackbody are independent of the particular material. Physicists can study the properties of intensity versus wavelength at fixed temperatures.
  3. Black-body Radiation 10 9 7 6 5 4 2 483 nrn Visible peak — 6000 K 580 nm 5000 K 4000 K r 724 nm aooo K 500 too 1000 1500 2000 966 nrn (IR) Wavelength {nm) UV 2.9 x 10-3 m 2500 3
  4. vs Temperature peak 2.9 x 10-3 m peak — 3100K (body temp) 58000K (Sun's surface) 2.9 x 10-3 m =9x10-6m 3100 infrared light visible light 2.9 x 10-3 m =O.5x10-6m 58000
  5. "Room temperature" radiation 750 K 500 K 350 K Wavelength X[vrn]
  6. Wien's Displacement Law (R, T) is the total power radiated per The intensity unit area per unit wavelength at a given temperature. Wien's displacement law: The maximum of the distribution shifts to smaller wavelengths as the temperature is increased. 2.0 1.5 1.0 0.5 0-0 1800 K 1500 K 1200 K 900 K 1 2 3 4 5 6 6 NV ave 1 g (x 1000 nm)
  7. Stefan-Boltzmann Law The total power radiated increases with the temperature: This is known as the Stefan-Boltzmann law, with the constant o experimentally measured to be 5.6705 x 10-8 W / (m2 1
  8. Rayleigh-Jeans Formula Lord Rayleigh (John Strutt) and James Jeans used the classical theories of electromagnetism and thermodynamics to show that the blackbody spectral distribution should be 2mckT 1200 K 81Tf2kT 3 C Rayleigh-Jeans forn•ula Experimental data 2000 4000 6000 8000 Wavelength (nm) It approaches the data at longer wavelengths, but it deviates badly at short wavelengths. This problem for small wavelengths became known as "the ultraviolet catastrophe" and was one of the outstanding exceptions that classical physics could not explain. 3 8M 3 1 -1 exp kT 3 8M 3 C 1 -1 exp(x) = 1 + x for very small x, i.e. when h -5 0, d. h. classical physics (also for f small and T large) 8
  9. Planck's solution EM energy cannot be radiated or absorbed in any arbitrary amounts, but only in discrete "quantum" amounts. The energy of a "quantum" depends on frequency as quantum h = 6.6 x 10-34Js "Planck's constant" 9
  10. Planck's quantum is small for "ordinary-sized" objects but large for atoms etc = hf E quant "ordinary" pendulum f=1Hz =6.6x10-34Jsx1Hz xeC =6.6x10-34J Hydrogen atom fæ 2x1014 Hz = hf E quant = 6.6x2)x 10-34+14J x 10-19J 10
  11. Typical energies in "ordinary" life Typical energy of a tot on a swing: Etot = mghmax = 20kgx10m/s2x1m max = 200 kgm2/s2 = 200 J much, much larger than =6.6x10-34J quant 11
  12. Typical electron KE in an atom Energy gained by an 1 "electron Volt" electron crossing a IV voltage difference Energy = q V lev = = 1.6x10-19 Joules similar = 1.3 x 10-19J E quant 12 for f z 2x1014 Hz
  13. Planck's Radiation Law Planck assumed that the radiation in the cavity was emitted (and absorbed) by some sort of "oscillators" that were contained in the walls. He used Boltzman's statistical methods to arrive at the following formula that fit the blackbody radiation data. 2mc2h 1 hc/1kT -1 Planck's radiation law Planck made two modifications to the classical theory: 1) 2) The oscillators (of electromagnetic origin) can only have certain discrete energies determined by En = nhf, where n is an integer, fis the frequency, and h is called Planck's constant. h = 6.6261 x 10-34 J.s. The oscillators can absorb or emit energy in discrete multiples of the fundamental quantum of energy given by 13
  14. Classical vs Quantum world In everyday life, quantum effects can be safely ignored This is because Planck's constant is so small At ato ic & subatomic scales, quantum effects are dominant & must be considered Laws of nature developed without consideration of quantum effects do not work for atoms 14
  15. Rayleigh-Jeans equation Consider the cavity as it emits blackbody radiation The power emitted from the blackbody is proportional to the radiation energy density in the cavity. One can define a spectral energy distribution such that is the fraction of energy per unit volume in the cavity with wavelengths in the range X to X + dX. Then, the power emitted at a given wavelength, R(X) u(X) u(X) may be calculated in a straightforward way from classical statistical physics. = (# modes in cavity in range dX) x (average energy of modes) # of modes in cavity in range dX, = 87tX-4 di Average energy per mode is kBT, according to kinetic theory u(X) = kBTn(X) = 8TtkBTX—4 15
  16. Wien, Rayleigh-Jeans and Planck distributions um (1) = c 81TkBT e-ß/ZT 14 1 e-23 1 e-26 1 e-27 1 e.29 15 Rayleigh-Jeans Wien Planck Wilhelm Carl Werner Otto Fritz Franz Wien 81Thc 15 (ehc/1kBT — 1) le 08 Frequency [Hz] 1 e+09 16
  17. The ultraviolet catastrophe There are serious flaws in the reasoning by Rayleigh and Jeans Furthermore, the result does not agree with experiment Even worse, it predicts an infinite energy density as 0! (This was termed the ultraviolet catastrophe at the time by Paul Ehrenfest) Rayleigh-Jeans law Planck's law 2000 4000 6000 X, nm Agreement between theory and experiment is only to be found at very long wavelengths. The problem is that statistics predict an infinite number of modes as classical kinetic theory ascribes an energy 1
  18. Planck's law (quantization of light energy) In fact, no classical physical law could have accounted for measured blackbody spectra Max Planck, and others, had no way of knowing whether the calculation of the number of modes in the cavity, or the average energy per mode (i.e. kinetic theory), was the problem. It turned out to be the latter. Planck found an empirical formula that fit the data, and then made appropriate changes to the classical calculation so as to obtain the desired result, which was non-classical. The problem is clearly connected with u(X) -+ T, as -+ 0 The problem boils down to the fact that there is no connection between the energy and the frequency of an oscillator in classical physics, i,e, there exists a continuum of energy states that are available for a harmonic oscillator of any given frequency. Classically, one can think of such an oscillator as performing larger and larger amplitude oscillations as its energy increases. 18
  19. Maxwell-Boltzmann statistics Define an energy distribution function f (E) = f (E) Then, This is simply the result that Rayleigh and others used, i.e. the average energy of a classical harmonic oscillator is kBT, regardless of its frequency. Planck postulated that the energies of harmonic oscillators could only take on discrete values equal to multiples of a fundamental energy e = hf, where fis the frequency of the harmonic oscillator, i.e. 0, e, 28, 38, etc Then, En = ne = nhf Here, h is a fundamental constant, now known as Planck's constant. Although Planck knew of no physical reason for doing this, he is credited with the birth of quantum mechanics. 19
  20. The new quantum statistics fn = Aexp(-En /kBT) = Aexp(-nhf /kBT) Replace the continuous integrals with a discrete sums: Efn = AEexp(-nhf /kBT) = = = x /kBT) n=0 Solving these equations together, one obtains: hc/ exp(8/kBT) —1 exp(hf exp(hc/1kBT) -1 Multiplying by D(X), to give —5 hc1 bl(/l) exp (hc / 1kBT) — 1 This is Planck's law 20
  21. The results of Planck's law Note: the denominator tends to infinity faster than the numerator (X- 5), thus resolving the infrared catastrophe, i.e. u(X) + O as —5 0, hc Note also: for very large X: exp (hc/ 1kBT) —l 1kBT From a fit between Planck's law and experimental data, one obtains Planck's constant to be: h = 6.626 x 10-34 J.s Planck's restriction of the available energies for radiation gets around the ultraviolet catastrophe in the following way: the short wavelength/high frequency modes are now limited in the energy they can have to either zero, or E 2 hf; in the calculation of the average energy, these modes with high energy are cut off by the Boltzmann factor exp(—E/kBT), i,e, these modes are rarely excited and, therefore, contribute nothing to the average energy in the limit --5 0, 21
  22. Photoelectric Effect Methods of electron emission: Thermionic emission: Application of heat allows electrons to gain enough energy to escape. Secondary emission: The electron gains enough energy by transfer from another high-speed particle that strikes the material from outside. Field emission: A strong external electric field pulls the electron out of the material. Incident light (electromagnetic radiation) shining on the materPhotoeIectric effect: ial transfers energy to the electrons, allowing them to escape. Electromagnetic radiation interacts with electrons within metals and gives the electrons increased kinetic energy. Light can give electrons enough extra kinetic energy to allow them to escape. We call the ejected electrons photoelectrons. 22

Need a Tutor or Coaching Class?

Post an enquiry and get instant responses from qualified and experienced tutors.

Post Requirement

Related PPTs

Upload PPTs

If you have your own PowerPoint Presentations which you think can benefit others, please upload on LearnPick. For each approved PPT you will get 100 Credit Points and 100 Activity Score which will increase your profile visibility.

Upload Now