{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1071_PartUniversity Physics Solution

1071_PartUniversity Physics Solution - Photons Electrons...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Photons, Electrons, and Atoms 38-15 38.67. I DENTIFY : Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the Stefan- Boltzmann law apply to its radiation. S ET U P : Wien’s displacement law is 3 peak 2.90 10 m K T λ × = , and the Stefan-Boltzmann law says that the intensity of the radiation is I = σ T 4 , so the total radiated power is P = σ AT 4 . E XECUTE : (a) First use Wien’s law to find the peak wavelength: λ m = (2.90 × 10 –3 m K )/(3000 K) = 9.667 × 10 –7 m Call N the number osf photons/second radiated. N × (energy per photon) = IA = σ AT 4 . N ( hc/ λ m ) = σ AT 4 . 4 m AT N hc λ σ = . 7 8 2 4 8 2 4 34 8 (9.667 10 m)(5.67 10 W/m K )(4 )(600 6.96 10 m) (3000 K) (6.626 10 J s)(3.00 10 m/s) N π × × × × = × × . N = 5 × 10 49 photons/s. (b) 2 4 4 2 4 B B B B B B S 4 2 4 S S S S S S S 4 600 3000 K 4 5800 K I A A T R T R I A A T R T R σ π σ π = = = ⎟ ⎜ = 3 × 10 4 E VALUATE : Betelgeuse radiates 30,000 times as much energy per second as does our sun! 38.68. I DENTIFY : The blackbody radiates heat into the water, but the water also radiates heat back into the blackbody. The net heat entering the water causes evaporation. Wien’s law tells us the peak wavelength radiated, but a thermophile in the water measures the wavelength and frequency of the light in the water. S ET U P : By the Stefan-Boltzman law, the net power radiated by the blackbody is ( ) 4 4 sphere water dQ A T T dt σ = . Since this heat evaporates water, the rate at which water evaporates is v dQ dm L dt dt = . Wien’s displacement law is 3 m 2.90 10 m K T λ × = , and the wavelength in the water is λ w = λ 0 / n . E XECUTE : (a) The net radiated heat is ( ) 4 4 sphere water dQ A T T dt σ = and the evaporation rate is v dQ dm L dt dt = , where dm is the mass of water that evaporates in time dt . Equating these two rates gives ( ) 4 4 v sphere water dm L A T T dt σ = . ( )( ) 2 4 4 sphere water v 4 R T T dm dt L σ π = . ( ) ( ) 8 2 4 2 4 4 4 3 5.67 10 W/m K 4 (0.120 m) (498 K) (373 K) 1.92 10 kg/s 0.193 g/s 2256 10 J/Kg dm dt π × = = × = × (b) (i) Wien’s law gives λ m = (0.00290 m K )/(498 K) = 5.82 × 10 –6 m But this would be the wavelength in vacuum. In the water the thermophile organism would measure λ w = λ 0 / n = (5.82 × 10 –6 m)/1.333 = 4.37 × 10 –6 m = 4.37 μm (ii) The frequency is the same as if the wave were in air, so f = c/ λ 0 = (3.00 × 10 8 m/s)/(5.82 × 10 –6 m) = 5.15 × 10 13 Hz E VALUATE : An alternative way is to use the quantities in the water: 0 / / c n f n λ = = c/ λ 0 , which gives the same answer for the frequency. An organism in the water would measure the light coming to it through the water, so the wavelength it would measure would be reduced by a factor of 1/ n . 38.69. I DENTIFY : The energy of the peak-intensity photons must be equal to the energy difference between the n = 1 and the n = 4 states. Wien’s law allows us to calculate what the temperature of the blackbody must be for it to radiate with its peak intensity at this wavelength.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern