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**Unformatted text preview: **PHY251_M2_S01_solutions Name: PHYSICS 251 - Final Exam Wednesday, May 16, 2001 Show all work for full credit! I. Short answer problems (120 points) 1. Calculate the momentum p and the wavelength l of: (Use eV/ c and nm units!) i. a photon of 10 eV energy, Solution: p = E / c =10 eV/ c ; l = h / p =1240 eV/ c nm/10 eV/ c = 124 nm. ii. an electron of 10 eV energy, Solution: p = √ (2 mE ) = √ (2×0.511×10- 6 ×10) = 3.2 keV/ c ; l = h / p =1240 eV/ c nm/3.2 keV/ c = 0.388 nm. iii. an electron of 10 GeV. Solution: p = E / c =10 GeV/ c ; l = h / p =1240 eV/ c nm/10 GeV/ c = 124×10- 9 nm = 0.124 fm. 2. A laser emits a beam of 10 17 photons per second, of wavelength 310 nm, which falls on a piece of blackened foil (mass 20 mg). Assume the photons are fully absorbed. i. Calculate the power P of the laser beam. Solution: E = hc / l = 4.0 eV; P = NE = 10 17 ×4.0 eV/s = 64 mW ii. Calculate the initial acceleration of the foil. Solution: p = E / c ; a = F / m = Ndp /( mdt ) = NdE /( mcdt ) = P /( mc ) = 64×10- 3 Nm/s /(2×10- 5 kg ×3×10 8 m/s) = 1.07×10- 5 m/s². 3. Four indistinguishable particles (with masses m = 5.0×10 5 eV/ c ²) are inside a cubical box, with sides equal to 1.0 nm. There is no magnetic field. i. If the particles have spin s = 0, calculate the ground state energy of the total system. Solution: The lowest energy level is at E 111 = 3 h ² / 8 mL ² = 3×0.384 eV = 1.15 eV. The next level is E 211 = E 121 = E 112 = 6 h ² / 8 mL ² = 6×0.384 eV = 2.30 eV. In the ground state, all four bosons are in the first level: E tot = 4× E 111 = 4.61 eV. ii. If the particles have spin s = 1 / 2 , calculate the ground state energy of the total system, and list the possible spin orientations of the particles. Solution: Only two fermions may occupy the lowest energy level with opposite spins. The other two, with opposing spins, must go into the first excited state; thus: E tot = 2×1.15 eV + 2×2.30 eV = 6.91 eV. iii. If the particles have spin s = 3 / 2 , calculate the ground state energy of the total system, and list the possible spin orientations of the particles. Solution: Now all fermions can go into the ground state, differentiated by their m s quantum number: m s = + 3 / 2 , + 1 / 2 , - 1 / 2 , - 3 / 2 . The total ground state energy is as in (i): 4.61 eV. 4. Compton scattering of 12.4 keV X-rays: i. Under what condition is Compton scattering observed? Solution: It assumes that the scattering is off quasi-free electrons! ii. Calculate the wavelength of X-rays Compton-scattered at a 60 degree angle with respect to the incident X-rays. Solution: l = 1240 eV·nm /12,400 eV = 0.1 nm; l ' = l- h / mc (1 - 1 / 2 ) = 0.100 + 1240/0.511×10- 6 ×½ = 0.1012 nm. ...

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