hw1_Solutions - Physics 226 Assignment#1 Solutions April 8...

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Physics 226 Assignment #1 Solutions April 8, 2015 1. (a) Room temperature is about 70 F 21 C 294 K. So kT 294 K × 8 . 6 × 10 - 5 eV/K 0 . 025 eV = 1/40 eV . Multiplying by 1 . 6 × 10 - 19 J/eV gives kT 4 × 10 - 21 J . (b) Kinetic energy (for a molecule moving slowly compared to c ) equals 1 2 mv 2 . If this equals 3 2 kT , then v 2 = 3 kT/m . Using a binomial expansion, the Lorentz factor γ = (1 - v 2 /c 2 ) - 1 / 2 1 + 1 2 v 2 /c 2 when v c . Hence, γ - 1 = 1 2 v 2 /c 2 = 3 2 kT/ ( mc 2 ). Air is predominantly (80%) composed of nitrogen molecules, with atomic weight 28. The mass m of a nitrogen molecule, in kilograms, is thus 28 amu × 1 . 66 × 10 - 27 kg/amu 4 . 6 × 10 - 26 kg. Hence, mc 2 4 . 6 × 10 - 26 kg × (3 × 10 8 m/s) 2 4 . 1 × 10 - 9 J. So finally, γ - 1 = 3 2 kT/ ( mc 2 ) (6 × 10 - 21 J) / (4 × 10 - 9 J) = 1 . 5 × 10 - 12 . 2. (a) The probability P ( t ) that a pion has not decayed in time t can be written in terms of the half-life t 1 / 2 as P ( t ) = 2 - t/t 1 / 2 , or in terms of the lifetime τ as P ( t ) = e - t/τ . Equate the (natural) logarithm of both forms: ln P ( t ) = - t/τ = ln(2 - t/t 1 / 2 ) = - ( t/t 1 / 2 ) ln 2. Therefore, 1 = (ln 2) /t 1 / 2 or τ = t 1 / 2 / (ln 2) 1 . 44 t 1 / 2 . (b) We are told that 2/3 of the pions survive to reach the detector a distance L 32 m away from the interaction point. So the time t taken by the pions to travel 32 meters — as measured by a clock comoving with the pions — must be such that 2 / 3 = P ( t ) = 2 - t/t 1 / 2 . But, because the pions are moving in the lab frame, a clock moving with the pions will appear to run slow, by a factor of γ , when viewed in the lab frame. In other words, when the comoving clock
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