Physics 3223 Solution to Assignment #5

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Hester Greene
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1 Physics 3223 Solution to Assignment #5 October 20, From the Heisenberg uncertainty principle, x p h, we estimate that a typical kinetic energy of an electron in an atom would be K ( p)2 2m h 2 c 2 2mc 2 ( x) 2 (200 MeV fm) 2 2(0.511 MeV)(0.1 nm) 2 ( ) 2 MeV 4 ev, which indeed is the order of observed electrons emitted from atomic transitions, so it is indeed possible for electrons to exist within atoms. 6.6 A 5 MeV alpha particle scatters off a gold nucleus at a 90 scattering angle. (a) The impact parameter is given by b zz e 2 cot θ 2K 4πɛ 0 2. (1) Putting in the numbers, since e 2 /4πɛ 0 α hc, b MeV fm cot fm. 5 MeV 137 1

2 (b) To find the distance of closest approach, we must solve the quadratic equation (6.21): 0 K + Kb2 r 2 m + zzα hc, (2) but because of the impact parameter equation (1) for cot θ/2 1, so the quadratic equation (2) reduces to 2Kb zzα hc, (3) 0 K + Kb2 r 2 m + 2Kb 0. Factoring K out, we can write this as or ( ) 2 b , b Thus, b fm fm. (c) At the minimum distance, the potential energy is, because of Eq. (3) U m zzα hc 2K b 2(5 MeV) MeV. The kinetic energy at the minimum distance is K m K b2 r 2 m 5 MeV ( ) MeV If if were not for round-off error, K m + U m K 5 MeV. 2

3 6.7 For a head-on collision, b 0, K 1 zze 2 4πɛ 0 d 2(79) 137 zzα hc d 200 Mev fm 7 fm 33 MeV The formula for the probability of scattering alpha particles into an area element r 2 dω is dn r 2 dω nt ( ) zz 2 α 2 h 2 c 2 1 4r 2 2K sin 2 (4) θ/2 Here the kinetic energy is K 6 MeV, the number density of gold atoms in the foil was calculated on page 181 of Krane to be n ρn z M m 3, the thickness of the foil is t m, z 2 and Z 79, the scattering angle is θ 30, and the distance to the detector is r 12 cm. To get the rate of scattering, we multiply Eq. (4) by the incident flux of alpha particles, F /s and multiply by the area of the detector, A π(0.5 cm) 2. The result of all this is the desired rate: Rate (3 107 /s)( m)( m 3 ) (79) 2 4(0.12 m) 2 (137) 2 ( ) Mev fm π( m) 2 6 MeV sin /s The formula for the energy for a photon emited when an electron makes a transition from a state from a state which quantum number n 1 to one with quantum number n 2 is ) hν 1 ( 1 2 mc2 α 2 1 n 2 2 n 2 1 For a given final state n 2, the largest frequency emited occurs when n 1, so ν limit α2 2 3 mc 2. hn 2 2.

4 The corresponding wavelength (the shortest wavelength of the series) is λ limit c ν limit 4π hcn2 2 α 2 mc 2. In class, we worked out the numbers for the Balmer limit, that is, for n 2 2: λ limit, Because the limiting wavelength is proportional to the square of n 2, n 1 : λ limit,1 λ limit,2 1 2 n 3 : λ limit,2 λ limit, The radius of the nth Bohr orbit is r n a 0 n 2, nm, nm. 22 so from the angulaomentum quantization condition we have But a 0 h/αmc, so mvr n h v n n h mr n v n h αmc mn h n h ma 0 n 2. αc n. For a hydrogenic atom, with nuclear charge Ze, we simply replace everywhere α Zα, so v n Zαc n By the substitution mentioned just above, for a hydrogenic atom with nuclear charge Ze, the energy of the nth state is E n Z2 α ev 2 mc2 Z2. n2 n 2 The ionization energy, the energy required to remove an electron from the atom, is just the negative of this. 4

5 (a) Z 1, n 3 : E ion 13.6 ev 1.51 ev. 9 (b) Z 2, n 2 : E ion ev 13.6 ev. 4 (c) Z 3, n 4 : E ion ev 7.65 ev (a) For a muonic atom, the energy of the nth state is E n 1 2 α2 m µ c 2 1 n 2, and so the limiting wavelength is for the n 2 1 series. (λ limit ) µ 4π hcn2 2 α 2 m µ c 2 (λ limit) e (λ limit) e nm 207 ( ) me m µ 0.44 nm, (b) Because the nucleus is not stationary, there is a center of mass correction: λ λ(1 + m/m), where M is the mass of the nucleus. Since the electron is only about 1/2000 of the mass of the proton, this is a very small correction for hydrogen, but for a muonic atom, since the muon is 207 times heavier than the electron, this correction is significant. (λ limit ) µ ( 1 + ) (λ limit ) 938 µ nm 0.49 nm. 5

6 6.37 If a whole number of wavelengths fit on the circumference of the orbit, so 2πr nλ n h p, L pr mvr n h 2π n h, which is exactly Bohr s quantization condition. 6

Chapter 1 The Bohr Atom

Chapter 1 The Bohr Atom 1 Introduction Niels Bohr was a Danish physicist who made a fundamental contribution to our understanding of atomic structure and quantum mechanics. He made the first successful