# ps11sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01T Fall Term 2004 Problem Set 11: Angular Momentum, Rotation and Translation Solutions Problem 1 : Bohr hydrogen atom The Bohr hydrogen atom models the atom as an electron orbiting the proton under the influence of an electric force producing uniform circular motion with radius a 0 . The mass of the electron is × m = 9.1 10 31 k g ; the electric charge is e = 1.6 × 10 19 C ; the Planck constant is e × h = 6.63 10 34 J s , and the magnitude of the electric force is given by Coulomb’s Law r 2 2 F = ke 2 r where k = 9.0 × 10 9 N m C 2 . The angular momentum is quantized according to 2 π Ln h = where n = 1, 2,. .. a) Write down the equation that arises from the application of Newton’s 2nd Law to the electron. Since the electron is much less massive than the proton the reduced mass is the mass of the electron. Te radial component of Newton’s Second Law is then 2 ke 2 2 = m v . e a 0 a 0 You can solve this equation for the velocity in terms of the radius of the electron orbit, ke 2 v = ma 0 e b) What is the angular momentum of the electron about the center of the atom? h Ln = m a v = 2 e 0 c) Using your results from parts a) and b) derive an equation for the radius a 0 of the atom as , , , a function of ne h m e , and k . Using the result for the velocity of the orbit, the angular momentum equation becomes 1

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2 h ke a 0 n = m . e 2 π m e Square this equation and solve for the radius a 0 of the atom 2 2 h 2 ke a 0 2 n = m 2 e 4 m e Thus 22 nh = a 0 4 2 mke 2 e , , , d) What is the energy E for the atom? Express your answer in terms of ne h m e , and k . n Using the value for the velocity of the orbit, the energy is given by 1 2 ke 2 1 ke 2 ke 2 ke 2 E = m v = m = . e e e a 0 2 a 0 2 a 0 2 m a 0 Now use the value for the radius of the orbit 2 2 4 1 ke 2 ke 2 2 mk e e E =− 22 . 2 2 2 a 0 2 n h / 4 m k e e Set 2 2 × 31 9 2 -2 2 19 4 2 2 4 ( 9.11 10 kg )( 8.99 × 10 N m C ) ( 1.60 × 10 C ) e A = = 2 × 34 2 h 2 ( 6.63 10 Js e c ) 18 A = ( 2.17 × 10 J ) 1ev = 13.6 ev . 1.60 × 10 19 J Then the energy is 2 2 4 ke 2 2 πµ k e 1 A E n 2 4 ke 2 h 2 2 2 2 n When n = 1 , the constant A = 13.6 ev = -E 1 is minus the ground state energy. 2
e) A hydrogen atom emits a photon which arise from an energy transition from the n = 3 to n = 1 energy level. Calculate the frequency of the light emitted f =−∆ Eh =−( E E 3 ) h . 1 Using our result for the energy levels, we have that A 1 (8)(13.6 ev)(1.6 × 10 19 J ev f = 1 = ( 9 ) ( 6 . 6 3 1 0 34 J s ) -1 ) = 3 . 4 6 × 1 0 15 J h 9 × f) What is the wavelength of the emitted light from part e)?

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ps11sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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