# Quantum Mechanics-solutions - hatch(heh595 Quantum...

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hatch (heh595) – Quantum Mechanics – yeazell – (55740)1Thisprint-outshouldhave11questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0pointsThe “seeing” ability, or resolution, of radia-tion is determined by its wavelength.Themassof anelectronsis9.10939×1031kg and Planck’s constant is 6.62607×1034J s.If the size of an atom is of the order of1.12 nm, how fast must an electron travel tohave a wavelength small enough to “see” anatom?Correct answer: 6.49455×105m/s.Explanation:Let :m= 9.10939×1031kg,λ= 1.12 nm,andh= 6.62607×1034J s.p=hλ=m vv=hm λ=6.62607×1034J s(9.10939×1031kg) (1.12 nm)=6.49455×105m/s.00210.0pointsA beam of electrons, each with the same ki-netic energy, illuminates a pair of slits sepa-rated by a distance of 58 nm. The beam formsbright and dark fringes on a screen located adistance 1.2 m beyond the two slits. The ar-rangement is otherwise identical to that usedin the optical two-slit interference experiment.The bright fringes are found to be separatedby a distance of 0.2 mm.What is the kinetic energy of the elec-trons in the beam?Planck’s constant is6.63×1034J·s.Correct answer: 16.1363 keV.Δy= 0.2 mm = 2×105m.r2r1yLdS1S2θ= tan1parenleftBigyLparenrightBigviewingscreenδdsinθr2-r1POnegationslashS2Q S190QK=p22msinθ=ΔyL,sodsinθ=n λdΔyL=n λλ=dΔyn L.The momentum isp=hλ=n L hdΔy,so the kinetic energy isK=p22m=n2L2h22m d2y)2=(1)2(1.2 m)22 (9.11×1031kg) (5.8×108m)2×(6.63×1034J·s)2(2×105m)2·eV1.6×1019J×1 keV1000 eV=16.1363 keV.
Explanation:Let :n= 1,L= 1.2 m,h= 6.63×1034J·s,m= 9.11×1031kg,d= 58 nm = 5.8×108m,and
hatch (heh595) – Quantum Mechanics – yeazell – (55740)200310.0pointsA particle is in the ground state of a box oflengthL.The wave function for a particle in theground state isφ(x) =
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