The Configuration of the Atom: Rutherford s Model

Transcription

1 CHAPTR 2 The Configuration of the Atom: Rutherford s Model Problem 2.2. (a) When α particles with kinetic energy of 5.00 MeV are scattered at 90 by gold nuclei, what is the impact parameter? (b) If the thickness of a gold foil is 1.0 µm, in what percentage of cases will the incident α particles be scattered at angles larger than 90 (this is called back-scattering). Solution 2.2. (a) From q. (2.6), we have b a 2 cot θ 2 Z 1Z 2 e 2 2 cot θ MeV fm cot fm MeV (b) For the 180 θ 90, the scattering probability η ntπb2 (θ 90 ) ρt A A πb 2 (θ 90 ) g cm cm g ( cm) %. Problem 2.3. When there occurs a head-on collision between a.5 MeV α particle and a gold nucleus, what is the distance of closest approach? What will it be if we replace the gold nucleus by a lithium nucleus? Solution 2.3. For gold nucleus, using q. (2.21b), r m a ( 1+ 1 ). 2 sin θ/2 7

2 8 Problems and Solutions in Modern Atomic and uclear Physics When θ 180, r m is the smallest distance. (r m ) min a Z 1Z 2 e MeV fm 51fm..5 MeV For a lithium nucleus, the smallest distance is (r m ) min a Z 1Z 2 e MeV fm fm..5 MeV 7 Problem 2.5. A narrow beam of protons with a kinetic energy of 1.0 MeV impinge perpendicularly on a gold foil of mass-thickness 1.5 mg/cm 2.A counter counts the protons scattered at an angle of 60. The window of the counter has an area of 1.5 cm 2, and the distance from the scattering region in the foil is 10 cm. The window faces and is at right angles to the protons that fall on it. What is the ratio between the numbers of protons that enter the window and that impinge on the foil? Solution 2.5. The ratio is η ρt ntσ(θ) Ω A A sin θ 2 S r 2, where a (Z 1 Z 2 e 2 )/ / fm. Hence we have η g cm 2 ( g 1 13 ) 2 cm sin Problem 2.7. A narrow and homogeneous α particle beam impinges perpendicularly on a tantalum foil of thickness 2.0 mg/cm 2. The ratio between the numbers of scattered and incident particles that are scattered at an angle θ>20 is Calculate the differential cross-section of a tantalum nucleus corresponding to the scattering angle θ 60. Solution 2.7. Since the scattering probability we have η sin θ 2 ntπb2 (θ 0 )ntπ a2 cot2 θ 0 2, a 2 η θ πnt tan2 0 2, σ c (θ) ηa tan2 θ0 2 (θ 0 20 ) πρt A sin θ 2 (θ 60 )

3 The Configuration of the Atom: Rutherford s Model π g/cm g tan sin cm 2 /sr 2 (b/sr). Problem 2.9. A narrow beam of protons with a kinetic energy of 1.0 MeV impinges perpendicularly upon a gold foil of mass thickness 1.5 mg/cm 2. Suppose the foil contains 30% silver. Calculate the relative number of particles that have a scattering angle larger than 30. Solution 2.9. The scattering probability is η ntπ a2 θ cot2 2 πρ At A (Z 1Z 2 e 2 ) 2 2 cot 2 θ 2 πρt (Z 1 e 2 ) 2 Z2 2 θ A 2 A cot2 2 where Z 1 1,ρt g cm 2, as the foil contains two kinds of atoms, Z2/A 2 should be replaced by ( / /108). Then we obtain η g cm g 1 ( MeV cm) MeV ( ) cot 2 15 Problem Assume Thomson s model is right: that the positive charge in an atom is uniformly distributed over the entire atom. If an α particle has an energy of 5.0 MeV and the radius of a gold atom is 1 Å, show that the largest angle of deflection of the α particle scattered from a gold atom would be about 10 rad (electrons are neglected). Solution The largest angle of deflection of the α particle is θ p p p F t Z 1Z 2 e 2 R 2 2R v 2Z 1Z 2 e 2 Rv where R is the radius of gold nucleus, thus θ 2Z 1Z 2 e 2 mv 2 R Z 1Z 2 e 2 R MeV fm 5.0 MeV fm.6 10 rad. Problem If α particles with kinetic energy up to 7.7 MeV energy are scattered by a gold foil and the Rutherford scattering formula is still correct, estimate the size of the gold nucleus.

4 10 Problems and Solutions in Modern Atomic and uclear Physics Solution The size of gold nucleus is approximately equal to the distance of closest approach (r m ) min. (r m ) min a Z 1Z 2 e MeV fm 7.7 MeV 30fm. Problem A narrow proton beam of 2.0 MeV impinges perpendicularly on a gold foil of mass-thickness 1.5 mg/cm 2. The beam current is 10 9 amp. A counter with a window of.0 mm diameter is put 10 cm away from the foil. The window faces the gold foil and is at 160 with respect to the incoming proton beam. How many proton counts would be accepted by this detector within 10 min? Solution From q. (2.15), the proton count per second accepted by the detector is: where I e ntσ(θ) Ω 10 9 C/s C s 1 nt ρt A A g cm g cm 2 Ω s l (0.2)2 cm 2 (10) 2 cm sr, ( Z1 Z 2 e 2 ) 2 ( ) MeV fm 1 σ(θ) 2.0 MeV Thus sin θ cm 2 sr 1. sin s cm sr cm 2 sr s 1. So, the proton counts within 10 min are: Problem The thickness of a thin gold layer on a silicon backing can be measured by Rutherford backscattering (RBS). A 2.0 MeV proton beam from an accelerator impinges perpendicularly on a thin gold layer and a detector with 1.0 mm diameter is placed 5.0 cm away from the

5 The Configuration of the Atom: Rutherford s Model 11 sample. The scattered protons at 160 are recorded. When the incoming protons have been accumulated up to 100 µc, the scattered proton count is What is the thickness of this gold layer? Solution Since the scattered proton count is given by nt Ω, sin θ 2 we have the thickness of the gold layer t ( ) 2 1 θ a n Ω sin 2 where, Q/e C/( C) , n ρ A /A g cm g 1 / cm 3, Ω s/l 2 π ( ) 2 /( 5 2 ) sr, a Z 1 Z 2 e 2 / MeV fm/2mev cm. Thus we get t ( cm sin cm. ) cm Problem A proton with kinetic energy of 25 kev is scattered by a helium nucleus at rest. The helium recoils in the direction of 60 with respect to the incoming proton. What is the impact distance of the incoming proton? Solution The impact distance of the incoming proton is b a 2 cot θ 2 Z 1Z 2 e 2 ( ) M + m cot θ c 2 M 2. When a proton is elastically scattered by a helium nucleus at rest, the recoil angle ψ c of helium in the center of mass system is equal to two times the angle ψ L in the laboratory system, that is: ψ c 2ψ L 120. Thus we have the scattering angle of the incoming proton in the center of mass system θ C π ψ c 60 ( ) MeV fm +1 b cot fm cm MeV