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 4momentum 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 4momentum 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 4momentum 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 4momentum 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 wavelength) 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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 Fall '08
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 Physics, Energy, Momentum, Photon, Special Relativity, lab frame, pinC

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