Physics II PH 112
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This 4 page Class Notes was uploaded by Jalon Willms on Monday October 19, 2015. The Class Notes belongs to PH 112 at Rose-Hulman Institute of Technology taught by Staff in Fall. Since its upload, it has received 54 views. For similar materials see /class/225117/ph-112-rose-hulman-institute-of-technology in Physics 2 at Rose-Hulman Institute of Technology.
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Date Created: 10/19/15
Knight Sections 381 through 384 21105 MJM Name Box A 39blackbody39 An ideal blackbody absorbs all radiation which falls on it so it is in some sense ideally 39black absorbing Ideas about blackbody absorbers had been around from at least the early 19th century Blackbody radiation Experimenters used a small hole in the side of a large oven to approximate a blackbody radiation entering the hole would supposedly go in and bounce around and not come out They measured the radiation coming out of the hole nding out how much of each wavelength it contained It turns out that the hotter a blackbody is the more it radiates and the more short wavelengths it radiates rather than long wavelengths Planck and Blackbody Radiation In 1900 Max Planck was able to come up with a theory of the emitted blackbody radiation which matched the experimental blackbody radiation data Planck had to assume that the walls of the cavity could only exchange energy in multiples of the frequency For a frequency f Planck assumed the walls could emit or absorb hf 2hf 3hf etc of energy But they couldn39t deal in odd chunks like 127 hf The theory had a constant h now called Planck s constant in it which Planck found by fitting the blackbody data Einstein reasoned that the electromagnetic radiation itself could only take on values like hf 2hf 6 hf etc but not 1273 hf This meant that if you shined light of frequency f on an object energy could be absorbed in 39chunks or quanta which are multiples of hf If you absorbed 4 hf it meant you had absorbed 4 chunks of light each of energy hf Later these chunks of light would be called photons Knight picks up the story from when in about 1890 Hertz found that a negatively charged electroscope could be discharged by shining ultraviolet light on it If you shined red light it did not discharge the electroscope and not with yellow or green light either And not with infrared light What would happen to an uncharged electroscope One of our famous and greatest formulas is v f7 where v is phase velocity f is the frequency and 7 is the wavelength You probably know that c speed of light in vacuum 3 x 108 ms We will write down some typical wavelengths of light and find the frequency and the energy E hf of an individual 39chunk of light or photon Infrared light 7 1000 nm 10000 A 10 x 10396 m 1 pm f cx 3x 108 ms10396 m 3 x1014 Hz Planck s constant is h 663 x 103934 Js so E hf 199 x 103919 J The work to move an electron through a potential difference of 1 volt is called the 39electronvolt W qV 16 x1019 C 1v 16 x1019 VC 16 x1019 J 1electronvolt 1eV So the energy ofthe infrared photon of wavelength 1 u is E 19916 eV 124 eV Yellow light The wavelength here is around 580 nm or 580 x 10397 m Exercise Find the energy of one 39photon39 of yellow light Eyellow photon 342 x 1019 J Eyellow photon 214 8V eV Ultraviolet light The wavelength here might be 350 nm or 350 x 10397 In Find E for a UV photon EUV 567x 10191 EUV 354 eV Einstein reasoned that the electroscope and other metal surfaces like it would not give up an electron unless struck by a minimum energy which he called a 39work function39 If a surface has a work function of 50 eV and you send in light of energy 20 eV the metal absorbs the energy but does not emit a photon Even if you hit it with lots of 20 eV photons no electrons come out Exercise suppose you hit the surface with a 60 eV photon would electrons come out Y N For a metal with a 39work function39 of 200 eV which light above if any would emit electrons infrared light yellow light ultraviolet U IO light There is a short way to calculate the energy E hf hclt You can readily show that hc 12400 eVA 1240 eVnm Then dividing the wavelength in nm gives the answer in eV IR E 1240 eVnm 1000 nm 124 eV Exercise Find the energy of a photon of wavelength 200 nm E 620 eV The sketch shows one version of a photoelectric experiment battery Big arrow light in jected electrof ca ode work function W The light comes in on the cathode and will eject a photoelectron if its frequency is high enough Sometimes the electron will have a kinetic energy equal to the difference between the photon energy and the work function This is the maximum KE the electron could have KEmaxhfW Exercise Find the maximum KE in eV of an electron ejected from a surface where W 21 eV by a photon of wavelength 400 nm Km 1240400 21 10 eV Some of the ejected photoelectrons will make their way to the top electrode in the sketch and create a current This will happen unless they are repelled and stopped by negative charges on the top electrode from the battery The work done on the electrons moving through a potential difference of meery is Work quanery Exercise Find the work done in eV on an electron travelling from cathode t0 anode if the battery voltage is 17 v W qAV 17eV 27x 1019 Exercise Now suppose the electron was ejected from a metal of work function W 21 eV by a photon ofwavelength 500 nm a What is the maximum KB of such an electron Km 1240500 21 038 eV b Will this electron be able to make it from the cathode to the anode with a 17 v battery opposing it No It would need a Kgreater than 1 7 th0 make it c What is the 39stopping voltage for the electron in part b What is the least battery voltage opposing it which will stop the electrode from reaching the anode ForK 038 thhe stopping voltage is 038 v deBroglie Wavelength Light can behave like particles under certain circumstances In these situations its energy is given by E hf hch Light also carries linear momentum in addition to energy The linear momentum p is quite small as you can see from the relation E hf pc light p hchc 1W light The momentum of a chunk of light is its energy divided by 300 million so it s quite small The momentum p is also planck s constant divided by the wavelength plight hh This applies to light but does it apply to an electron or a neutron The idea of calculating the momentum of a particle from the formula pmvh7t seems stupid because the particle doesn t have a wavelength Or does it Actually the boxed formula just above was proposed by deBroglie in around 1923 and was found to be correct by experiments done in 1927 and 1928 Let s find the wavelen h of an electron accelerated from rest throu a otential difference of 20 v The work done is qV 20 eV This equals the change in KE 20 X 16 X 103919 J 12 mvz With m 91X103931 kg we find v 265 X 106 ms Then mv 241 X 103924 kgms Solving from boxed formula with h 663 X103934js we get A 274 x1010 m or A 274 A 0274 nm Exercise A proton m167 x 10 27 kg has a kinetic energy of 50 eV Find its wavelength 50x 16x 10191 12 167x 1027 kg v2 gt v 9 79x 10quot ms 1 663 x 10341441 67x 10 27 9 79x 104ms 406x 1039 m Exercise An electron is ejected from a metal of W 190 eV by a photon 0f 7 400 nm Assuming the electron to have the maximum possible KE determine the electron39s wavelength Km 31 19 12 eV 19x 10191 1 hVQmKW 663x103934591 x 1039 kgms 112x 10399 m 11 nm Exercise What is the longest wavelength of light which will eject electrons from a metal surface of work function W 25 eV hc ilong W gt Along hcW 124025 496 nm this is blue green light
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