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Unformatted text preview: PHY251_M2_S01_solutions Name: PHYSICS 251 - Midterm II Wednesday, April 18, 2001 Show all work for full credit! 1. Short answer problems (100 points) a. An unstable system emits a photon of l = 121.50 nm which has an uncertainty in its wavelength of Dl = 0.24 nm. i. Calculate the corresponding uncertainty in the energy DE of the photon. E = hc / l ;= 1240 eVnm / 121.50 nm = 10.206 eV DE = ( E / l ) Dl = ( hc / l ) Dl = E ( Dl / l ) = 1240 eVnm 0.24 nm / (121.50 nm) = 0.020 eV. ii. Calculate the lifetime Dt of the unstable system. DE Dt h / 2 ; thus: Dt h / 2 DE = 197 eVnm/ c / 20.4 eV = 3.210- 17 s. b. the Pauli Principle i. State the Pauli principle. The Pauli principle states that no two identical fermions (half-integer spin particles) can co-exist in the same location. Identical: having the same set of quantum numbers; same location: when their wavefunctions significantly overlap. ii. Define: (a) fermions, (b) bosons. a. Fermions: half-integer spin particles that follow Fermi-Dirac statistics. They have fully anti-symmetric wavefunctions; as a consequence they obey the Pauli principle. b. Bosons: integer spin particles that obey Bose-Einstein statistics. They have fully symmetric wavefunctions. c. Consider an electron in a one-dimensional infinite potential well of width L = 0.123 nm. i. Calculate the ground state energy, and the energy spectrum E n (all values in eV!) E n = n ( h / 8 mL ) = n (1240 eVnm)/(8 0.51110 6 eV 0.123 nm) = n (24.9 eV); thus the ground state energy is 24.9 eV. file:///C|/Documents%20and%20Settings/Linda%20Grau...0HEP-H2/PHY251%20Sp01/PHY251_M2_S01_solutions.html (1 of 9) [2/4/2008 4:39:36 PM] PHY251_M2_S01_solutions ii. One may take the electron's position inside the well to be uncertain with Dx 2245 0.1 nm. Using the uncertainty principle, calculate the minimum kinetic energy (in eV!) the electron must have inside the well. Dp x Dx h / 2 ; thus: Dp x h / 2 Dx = 197 eVnm/ c / 0.2 nm = 1 keV/ c . The average momentum p = 0 (as many times right as left, averages to zero), and therefore Dp x = p - p = p . Thus K = p / 2 m (1 keV/ c )/(20.51110 6 eV/ c ) = 1.0 eV iii. Using the result from (i), calculate the ground state energy of the total well system with three electrons in it. Electrons are fermions: two, with opposite spins, can occupy the n =1 ground state; however the third electron has to go into the next higher state n =2. The total energy of the system is thus: E tot = 2 E 1 + 1 E 2 = 2 E 1 + 14 E 1 = 6 E 1 = 624.9 eV = 149 eV. iv. Using the result from (i), calculate what the the ground state energy of the total well system will be if the three electrons in (iii) were replaced with three bosons of equal mass....
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