hw1_Solutions

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern