UNIVERSITY PHYSICS UCI PKG
11th Edition
ISBN: 9781323575208
Author: YOUNG
Publisher: PEARSON C
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Chapter 39, Problem 39.38E
(a)
To determine
The wavelength corresponding to the peak of the planck distribution.and frequency corresponding to
(b)
To determine
The wavelength corresponding to the peak of the planck distribution.and frequency corresponding to
(c)
To determine
The wavelength corresponding to the peak of the planck distribution.and frequency corresponding to
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Problem-1:
An asteroid is hurtling toward earth at 150,000“. The temperature of the asteroid is about 100 K, meaning that its peak emission
is 2 = 29 µm. The speed of light is c =
3E[8].
a) What is the wavelength of light that we receive from the asteroid? (Answer: 2.89855E[-05] m)
8πhy3 dv
(a) Express the Planck radiation formula, Edv =
c3 ehv/kT-1'
terms of λ (and dλ), namely Edλ. (b) Solved=0 for λ = Amax
in
to determine the value of maxT in m-K, where max is the
wavelength at which the blackbody spectrum has its maximum value
at a given temperature T.
The energy emitted by a black body's surface per unit area at a particul ar wavelength can
be calcul ated using Planck's Radiation Law, which can be written as follows,
2nhc?
E(2, T) =
25. (ehc/kT -1)
where his Planck's constant 6.626 x 10-27 erg.s, c is the speed oflight, kis the Boltzmann
constant = 1.38 x 10-18 erg/K, Tis the temperature in Kelvins and A is the wavelength in
um.
If E is given in erg/um?, what are the units of the constant 2n, given that the equation is
valid and therefore dimensionally homogenous? If the speed of light is 3.00x10$ m/s, what
value and unit should be used for c in this equation to maintain dimensional homogeneity?
Note: erg is a unit of energy equal to 10-7 Joules and the whole expression
(ehc/AkT – 1) ends up dimensionless.
-
Chapter 39 Solutions
UNIVERSITY PHYSICS UCI PKG
Ch. 39.2 - Prob. 39.2TYUCh. 39.3 - Prob. 39.3TYUCh. 39.4 - Prob. 39.4TYUCh. 39.5 - Prob. 39.5TYUCh. 39.6 - Prob. 39.6TYUCh. 39 - Prob. 39.1DQCh. 39 - Prob. 39.2DQCh. 39 - Prob. 39.3DQCh. 39 - When an electron beam goes through a very small...Ch. 39 - Prob. 39.5DQ
Ch. 39 - Prob. 39.6DQCh. 39 - Prob. 39.7DQCh. 39 - Prob. 39.8DQCh. 39 - Prob. 39.9DQCh. 39 - Prob. 39.10DQCh. 39 - Prob. 39.11DQCh. 39 - Prob. 39.12DQCh. 39 - Prob. 39.13DQCh. 39 - Prob. 39.14DQCh. 39 - Prob. 39.15DQCh. 39 - Prob. 39.16DQCh. 39 - Prob. 39.17DQCh. 39 - Prob. 39.18DQCh. 39 - Prob. 39.19DQCh. 39 - Prob. 39.20DQCh. 39 - Prob. 39.21DQCh. 39 - When you check the air pressure in a tire, a...Ch. 39 - Prob. 39.1ECh. 39 - Prob. 39.2ECh. 39 - Prob. 39.3ECh. 39 - Prob. 39.4ECh. 39 - Prob. 39.5ECh. 39 - Prob. 39.6ECh. 39 - Prob. 39.7ECh. 39 - Prob. 39.8ECh. 39 - Prob. 39.9ECh. 39 - Prob. 39.10ECh. 39 - Prob. 39.11ECh. 39 - Prob. 39.12ECh. 39 - Prob. 39.13ECh. 39 - Prob. 39.14ECh. 39 - Prob. 39.15ECh. 39 - Prob. 39.16ECh. 39 - Prob. 39.17ECh. 39 - Prob. 39.18ECh. 39 - Prob. 39.19ECh. 39 - Prob. 39.20ECh. 39 - Prob. 39.21ECh. 39 - Prob. 39.22ECh. 39 - Prob. 39.23ECh. 39 - Prob. 39.24ECh. 39 - Prob. 39.25ECh. 39 - Prob. 39.26ECh. 39 - Prob. 39.27ECh. 39 - Prob. 39.28ECh. 39 - Prob. 39.29ECh. 39 - Prob. 39.30ECh. 39 - Prob. 39.31ECh. 39 - Prob. 39.32ECh. 39 - Prob. 39.33ECh. 39 - Prob. 39.34ECh. 39 - Prob. 39.35ECh. 39 - Prob. 39.36ECh. 39 - Prob. 39.37ECh. 39 - Prob. 39.38ECh. 39 - Prob. 39.39ECh. 39 - Prob. 39.40ECh. 39 - Prob. 39.41ECh. 39 - Prob. 39.42ECh. 39 - Prob. 39.43ECh. 39 - Prob. 39.44ECh. 39 - Prob. 39.45ECh. 39 - Prob. 39.46ECh. 39 - Prob. 39.47ECh. 39 - Prob. 39.48ECh. 39 - Prob. 39.49ECh. 39 - Prob. 39.50PCh. 39 - Prob. 39.51PCh. 39 - Prob. 39.52PCh. 39 - Prob. 39.53PCh. 39 - Prob. 39.54PCh. 39 - Prob. 39.55PCh. 39 - Prob. 39.56PCh. 39 - Prob. 39.57PCh. 39 - Prob. 39.58PCh. 39 - Prob. 39.59PCh. 39 - An Ideal Blackbody. A large cavity that has a very...Ch. 39 - Prob. 39.61PCh. 39 - Prob. 39.62PCh. 39 - Prob. 39.63PCh. 39 - Prob. 39.64PCh. 39 - Prob. 39.65PCh. 39 - Prob. 39.66PCh. 39 - Prob. 39.67PCh. 39 - Prob. 39.68PCh. 39 - Prob. 39.69PCh. 39 - Prob. 39.70PCh. 39 - Prob. 39.71PCh. 39 - Prob. 39.72PCh. 39 - Prob. 39.73PCh. 39 - Prob. 39.74PCh. 39 - Prob. 39.75PCh. 39 - Prob. 39.76PCh. 39 - Prob. 39.77PCh. 39 - Prob. 39.78PCh. 39 - Prob. 39.79PCh. 39 - Prob. 39.80PCh. 39 - A particle with mass m moves in a potential U(x) =...Ch. 39 - Prob. 39.82PCh. 39 - Prob. 39.83PCh. 39 - DATA In the crystallography lab where you work,...Ch. 39 - Prob. 39.85PCh. 39 - Prob. 39.86CPCh. 39 - Prob. 39.87CPCh. 39 - Prob. 39.88PPCh. 39 - Prob. 39.89PPCh. 39 - Prob. 39.90PPCh. 39 - Prob. 39.91PP
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- 2.3. Find the de Broglie wavelength of (a) an electron, and (b) a proton with speeds of 5 × 106 m/s and compare with the radius of the hydrogen atom, ao. Would either of these particles behave like a wave inside the H atom?arrow_forwardPlanck's principal assumption was that the energies of the electronic oscillators can have only the values &n=nhv and that Aε = hv. (a) Further assume that the number of oscillators with the energy En, Nn, is proportional to e-En/kT at the temperature T, namely Nn xe-En/kT, where N is the total number of oscillators. Show N that the average energy per oscillator is ɛ̃ = formula, Edv = formula as v → 0. ΣNnEn N c3 ehv/kT-1' = (b) As v→ 0, then Aɛ → 0 and & is essentially continuous. Hence, we should expect the non-classical Planck distribution to go over to the classical Rayleigh-Jeans distribution at low frequencies, where Ac→ 0. Show that the Planck radiation 8πhy3 dv reduces to the Rayleigh-Jeans hv ehv/kT-1arrow_forwardCalculate the wavelength of electrons accelerated through 1742 V. (electron mass m = 9.1 x 10-31 Kg, electronic charge e=1.6 x 10-19 C and the Planck constant h=6.6 x 10-34 Js) Round off the answer to 2 decimal places and write answers as power of 10.arrow_forward
- 3. The equation 1.45 in our textbook says that the de Broglie wavelength as a function of temperature is given by 1 Assuming the accepted accuracy for h, R, and m are very high, what is the V3MRT maximum uncertainty in the de Broglie wavelength if AT is the uncertainty in temperature?arrow_forwardDo it asaparrow_forwardThe root mean square speed of the hydrogen molecules at temperature t °C is given by 3x8.31 x (t+273) m 2 x 10-3 Calculate the de Broglie wavelength (in nanometers) of the hydrogen molecules at temperature 24 °C. The mass of the hydrogen molecule is 2 x 1.66 x 10-27 kg. Use two decimals in your answer.arrow_forward
- (b) Calculate the de Broglie wavelength of an electron having a mass of 9.11 x 10-31 kg and a charge of 1.602 x 10-19 J with a Kinetic energy of 110 eV. The value of the Planck’s constant is equal to 6.63 * 10-34 Js.arrow_forwardConsider a black body of surface area 22.0 cm² and temperature 5700 K. (a) How much power does it radiate? 131675.5 W (b) At what wavelength does it radiate most intensely? 508.421 nm (c) Find the spectral power per wavelength at this wavelength. Remember that the Planck intensity is "intensity per unit wavelength", with units of W/m³, and "power per unit wavelength" is equal to that intensity times the surface area, with units of W/m 131.5775 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. W/marrow_forwardPlanck’s constant has the value h = 6.626 × 10–34 joule-seconds (J-s), and the speed of light is c = 3 × 108 m/s. Using these values, calculate the wavelength carried by photons emitted with an energy of 1.1 × 10-19 J. Pick the closest value:arrow_forward
- Planck's radiation law can be written ux = 8лhc 1 25 eßhc/2-1 Show that the wavelength corresponding to the maximum energy density of the radiation fulfills the condition λmax T = . constant What is this constant? (This result is known as Wien's transition law.) Tip: you can solve the constant approximation by e.g. iterating an equation of the form Xn = 5 (1-e¯Xn-1) with a suitable initial value x1.arrow_forwardA blackbody (a hollow sphere whose inside is black) emits radiation when it is heated. The emittance (Mλ, W/m3), which is the power per unit area per wavelength, at a given temperature (T, K) and wavelength (λ, m) is given by the Planck distribution, where h is Planck's constant, c is the speed of light, and k is Boltzmann's constant. Determine the temperature in degrees Celsius at which a blackbody will emit light of wavelength 3.57 μm with an Mλ of 5.31×1010 W/m3. The power per unit area emitted can be determined by integrating Mλ between two wavelengths, λ1 and λ2. However, for narrow wavelength ranges (Δλ), the power emitted can be simply calculated as the product of Mλ and Δλ. power emitted=MλΔλ Using the conditions from the first part of the question, determine the power emitted per square meter (W/m2) between the wavelengths 3.56 μm and 3.58 μm.arrow_forwardA particular star has a radius of 8.46 ✕ 108 m. The peak intensity of the radiation it emits is at a wavelength of 679 nm. (a) What is the energy (in J) of a photon with this wavelength? answer in J (b) What is the star's surface temperature (in K)? (Round your answer to at least the nearest integer.) answer in K (c) At what rate (in W) is energy emitted from the star in the form of radiation? Assume the star is a blackbody, with emissivity e = 1. answer in W (d) Using the results from parts (a) and (c), estimate the rate (in photons/s) at which photons are emitted by the star. answer in photon/sarrow_forward
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