1A) How wide will the infinite 1-D square well be for electrons inthe lowest sta

Question

1A) How wide will the infinite 1-D square well be for electrons inthe lowest state to have a velocity of .4c? B) If 3 electrons are put into this well (and fall tolowest possible state), what is the kinetic energy of eachelectron? (ignore electrostatic interactions betweenelectrons) C) Find the three lowest photon energies that would beabsorbed by this system. B) If 3 electrons are put into this well (and fall tolowest possible state), what is the kinetic energy of eachelectron? (ignore electrostatic interactions betweenelectrons) C) Find the three lowest photon energies that would beabsorbed by this system.

Explanation / Answer

a) if velocity v = .4c = 1/(1-.4^2) = 1.091 momentum p = mv = (.511 Mev/c^2)(.4c)(1.091) = .223 MeV/c KE = (p^2c^2+m^2c^4) - mc^2 = (.223^2+.511^2) - .511= .047 MeV if i had used the non-relativistic formula, KE = (1/2)(.511 MeV/c^2)(.4c)^2 = .041 MeV ground state energy = E1 = h^2/(8mL^2) = .047 MeV L = h/(8*.511 MeV/c^2 * .047 MeV) = hc/(.438 MeV) = 1240MeV.fm / .438 MeV = 2829 fm b) lowest state takes 2 electrons (spin up and spin down) first excited state takes 3rd electron. c) the 3 lowest photon energies would be E2 - E1 = 3 E1 E3 - E1 = 8 E1 E3 - E2 = 5 E1

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