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LinSol - n = 1 and n = 2 we get E 1 =-k 2 Z 2 e 4 m π 2 ~...

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Problem 7 Part a) For circular motion we know that the acceleration is given by a 2 = v 2 r . The force responsible for the motion in this case is the Coulomb force between the electron and the Fe nucleus, which is given by: F = kZe 2 r 2 , where k = 1 4 π 0 = 8 . 99 × 10 9 Jm/C 2 and Z = 26 Setting this equal to the usual F = ma , gives us the velocity as follows: m π a = kZe 2 r 2 m π v 2 r = kZe 2 r 2 v = s kZe 2 m π r Part b) The angular momentum is given by L = m π vr , since for circular motion it is always true that ~v ⊥ ~ r . Then setting this equal to n ~ and substituting from part a), we get: m π vr = n ~ m 2 π kZe 2 m π r r 2 = n 2 ~ 2 r = n 2 ~ 2 kZe 2 m π The energy is found using the usual kinetic plus potential, where the Coulomb potential energy is U = - Ze 2 r : E = 1 2 m π v 2 - kZe 2 r E = 1 2 m π kZe 2 m π r - kZe 2 r E = - kZe 2 2 r , plugging in for r E = - k 2 Z 2 e 4 m π
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Unformatted text preview: n = 1 and n = 2 we get: E 1 =-k 2 Z 2 e 4 m π 2 ~ 2 =-(8 . 99 × 10 9 Jm/C 2 ) 2 (26 2 )(1 . 60 × 10-19 C ) 4 (273)(9 . 11 × 10-31 kg ) 2(1 . 05 × 10-34 J · s ) 2 =-4 . 04 × 10-13 J =-2 . 52 × 10 6 eV E 2 =-k 2 Z 2 e 4 m π 8 ~ 2 =-(8 . 99 × 10 9 Jm/C 2 ) 2 (26 2 )(1 . 60 × 10-19 C ) 4 (273)(9 . 11 × 10-31 kg ) 8(1 . 05 × 10-34 J · s ) 2 =-1 . 01 × 10 6 J =-6 . 31 × 10 5 eV Part c) Finally, the energy of the photon emitted upon transition from level 2 to level 1 is the difference between the energies of the two levels: E = E 2-E 1 = 1 . 89 × 10 6 eV 1...
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