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eqn1 - PHY 251 equation sheet for exam 1 version 0 = with =...

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PHY 251 equation sheet for exam 1 (September 22, 2009 version) ν = ν 0 γ [1 + ( v/c )cos θ ] with θ = 0 for emitter moving directly away. γ = 1 radicalBig 1 v 2 /c 2 p = γm 0 v , F = γm 0 a , F bardbl = γ 3 m 0 a . x 2 = γ ( x 1 vt 1 ) y 2 = y 1 z 2 = z 1 t 2 = γ parenleftbigg t 1 v c 2 x 1 parenrightbigg and v 2 ,x = v 1 ,x v 1 v v 1 ,x c 2 v 2 ,y = v 1 ,y γ bracketleftbigg 1 v v 1 ,x c 2 bracketrightbigg v 2 ,z = v 1 ,z γ bracketleftbigg 1 v v 1 ,x c 2 bracketrightbigg E = E 0 + E k = m 0 c 2 + ( γ 1) m 0 c 2 , E 2 = E 2 0 + p 2 c 2 . p x, 2 = γ parenleftBig p x, 1 v ( E/c 2 ) parenrightBig p y, 2 = p y, 1 p z, 2 = p z, 1 E 2 = γ ( E vp x ) . E = = hc/λ , 1 eV= 1 . 602 × 10 19 Joule, h = 6 . 62 × 10 34 J · sec, h = 4 . 14 × 10 15 eV · sec. hc = 1239 . 8 eV · nm. k B = 1 . 38 × 10 23 J/K, k B = 8 . 62 × 10 5 eV/K, c = 3 . 00 × 10 8 m/sec. u ν = 8 πhν 3 c 3 1 exp[ hν/kT ] 1 λ peak = hc/ (4 . 965 k B T ) E k = φ . λ = h/p , λ s λ 0 = h m e c (1 cos θ ) . Centripetal force to maintain circular motion: γmv 2 /r . Lorentz force:
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