WeekProb_Oct26Sol

WeekProb_Oct26Sol - Physics 7C Section 1 Midterm 2 Solution...

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Unformatted text preview: Physics 7C Section 1 Midterm 2 Solution : Problem 4 a) Wave length of initial photon (λ = 50 × 10−12 m) is related to its momentum and energy by hc E= (1) λ h (2) p= λ Thus 4-momentum of photon in lab frame is Eγ h pγ x − h λ pγ y = 0 0 pγ z c λ (3) Since electron is traveling at v = 0.98c in +ˆ direction in lab frame, to x boost into electron’s rest frame γ factor required is γ= 1− Similarly β is 1 0.98c 2 c = 5.0 (4) v = 0.98 (5) c I find energy and momentum of photon in electron’s rest frame by making Lorentz transformation of 4-momentum h Eγ ′ γ −γβ 0 0 λ c h p′ −γβ γ 0 0 −λ γx = (6) p′ 0 0 1 0 0 γy 0 0 01 0 p′γ z β= Solving this transformation I obtain ′ Eγ c p′γ x p′γ y p′γ z 9.9 h λ −9.9 h λ = 0 0 1 (7) So I found wavelength of initial photon in electron’s rest frame to be λ′ = λ = 5.1 × 10−12 m 9.9 (8) We derived formula for scattered photon’s wave length in electron’s rest frame h λ′f − λ′i = (1 − cos θ′ ) (9) me c Where θ′ is scattered angle of photon in electron’s rest frame. Since photon scattered at 180◦ in lab, it also did scatter 180◦ in electron’s rest frame. You can show this by arguing that scattering by 180◦ in lab frame means scattered photon only has px component of momentum in lab frame. As we saw earlier in how 4-momentum transforms under Lorentz transformation, scatterd photon only has p′x component in electron’s rest frame. Meaning photon scattered by 180◦ in both frame. Now using cos(180◦ ) = −1 we can solve for λ′f λ′f = λ′i + 2 λ′f h me c = 5.1 × 10−12 m + 2(2.43 × 10−12 m) (10) (11) (12) λ′f = 10 × 10−12 m Since we want to find λf in lab frame, we can transform photon’s 4-momentum by Eγ h γ γβ 0 0 λ′ f c h pγ x γβ γ 0 0 ′ λf (13) 0 = 0 0 1 0 0 0 0 01 0 0 Notice that direction of momentum is in +ˆ direction. Solving above equax tion I obtain Eγ 9.9 h λ′ f c pγ x 9.9 h′ λf (14) 0 = 0 0 0 Meaning λf = 2 λ′f 9.9 (15) Plugging in λ′f = 10 × 10−12 m λf = 1.0 × 10−12 m (16) b) Since we know initial and final energy of photon (we can obtain this from wave-length) and initial energy of electron, we can use conservation of energy to find final energy of electron. final final initial initial Eγ + Ee− = Eγ + Ee− (17) Plugging in initial Eγ = final Eγ = 25KeV λinitial γ hc = final = 1.2MeV λγ hc (18) (19) (20) initial = γme− c2 = 2.6MeV Ee− Final energy of electron is final Ee− = 1.4MeV (21) 3 Physics 7c Quiz 10 Solution Fall 2007 1. (a) 4 4 2 i. Ps = σTS As = σTS 4πRS . ii. The fraction that hits earth is the crossectional area of earth over the area of a sphere at the distance of earth’s orbit, so the incident power on earth is Pinc = PS ∗ 4 22 2 σTS πRS RE πRE = . 2 2 4πD D 4 4 2 iii. The total power leaving the earth is PE = σTE AE = σTE 4πRE . iv. Setting Pinc = PE and solving for TE , we find TE = TS RS ≈ 280K. 2D This is pretty close to 300K , but definitely not right, since it is barely above freezing, and as noted, a substantial amount of the sun’s light is reflected. (b) Using λm T = 2e898 × 106 nm · K , we find that λm,E ≈ 10µm and λm,S ≈ 500nm. (c) Now suppose that only the fraction f of the power leaving the earth escapes. Then 4 2 we adjust PE = f σTE 4πRE , and again equating this to Pinc and solving for f , we find T 4 R2 f = S4 S2 ≈ 0.76. 4TE D 2. (a) Conservation of momentum in the x−direction: pi = pf cos θ + pe cos φ. Conservation of momentum in the y −direction: pf sin θ = pe sin φ. Conservation of energy mc2 + pi c = pf c + Ee where Ee = p2 c2 − m2 c4 . e (b) Rewriting the conservation above formulas, we find pe cos φ = pi − pf cos θ, pe sin φ = pf sin θ, and Ee = mc2 + pi c − pf c. (c) Plugging the above into the equation of the rest mass of the electron, 2 2 (mc2 )2 = Ee − (pe c)2 = Ee − ((pe c cos φ)2 + (pe c sin φ)2 ) = c2 (mc + pi − pf )2 − c2 (pi − pf cos θ)2 − c2 (pf sin θ)2 m2 c2 = m2 c2 + 2mc(pi − pf ) + p2 − 2pi pf + p2 − p2 + 2pi pf cos θ − p2 cos2 θ − p2 sin2 θ i f i f f 0 = 2mc(pi − pf ) − 2pi pf (1 − cos θ), which we may rewrite as m(pi c − pf c) = pi pf (1 − cos θ). (d) Since λ = h , substitute pi = p mch h λi and pf = 1 1 − λi λf 1 h λf , so substituting this we find h2 (1 − cos θ), λi λf = or λf − λf = h (1 − cos θ). mc 2 ...
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