{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

final

# final - PHYSICS 2D PROF HIRSCH FINAL EXAM WINTER QUARTER...

This preview shows pages 1–3. Sign up to view the full content.

PHYSICS 2D FINAL EXAM WINTER QUARTER 2011 PROF. HIRSCH MARCH 14th, 2011 Formulas: Time dilation; Length contraction: Δ t = γ Δ t ' γ Δ t p ; L = L p / γ ; c = 3 × 10 8 m / s Lorentz transformation: x ' = γ ( x vt ) ; y '= y ; z '= z ; t ' = γ ( t vx / c 2 ) ; inverse : v - v Spacetime interval: ( Δ s ) 2 = ( c Δ t ) 2 -[ Δ x 2 + Δ y 2 + Δ z 2 ] ; γ =1/ 1 v 2 / c 2 Velocity transformation: u x ' = u x v 1 u x v / c 2 ; u y ' = u y γ (1 u x v / c 2 ) ; inverse : v - v Relativistic Doppler shift : f obs = f source 1 + v / c / 1 v / c (approaching) Momentum: r p = γ m r u ; Energy : E = γ mc 2 ; Kinetic energy: K = ( γ 1) mc 2 Rest energy : E 0 = mc 2 ; E = p 2 c 2 + m 2 c 4 Electron : m e = 0.511 MeV / c 2 Proton : m p = 938.26 MeV / c 2 Neutron : m n = 939.55 MeV / c 2 Atomic mass unit : 1 u = 931.5 MeV / c 2 ; electron volt : 1eV =1.6 × 10 -19 J Stefan's law : e tot = σ T 4 , e tot = power/unit area ; σ = 5.67 × 10 8 W / m 2 K 4 e tot = cU /4 , U = energy density = u ( λ , T ) d λ 0 ; Wien's law : λ m T = hc 4.96 k B Boltzmann distribution: P ( E ) = Ce - E /( k B T ) Planck's law : u λ ( λ , T ) = N λ ( λ ) × E ( λ , T ) = 8 π λ 4 × hc / λ e hc / λ k B T 1 ; N ( f ) = 8 π f 2 c 3 Photons: E = hf = pc ; f = c / λ ; hc =12,400 eV A ; k B = (1/11,600) eV / K Photoelectric effect : eV s = K max = hf φ , φ work function; Bragg equation : n λ = 2 d sin ϑ Compton scattering : λ '- λ = h m e c (1 cos θ ); h m e c = 0.0243 A ; Coulomb constant : ke 2 = 14.4 eV A Coulomb force : F = kq 1 q 2 r 2 ; Coulomb potential: V = kq r ; Coulomb energy : U = kq 1 q 2 r Force in electric and magnetic fields (Lorentz force): r F = q r E + q r v × r B Rutherford scattering: Δ n = C Z 2 K α 2 1 sin 4 ( φ /2) ; h c = 1,973 eV A Hydrogen spectrum: 1 λ mn = R ( 1 m 2 1 n 2 ) ; R = 1.097 × 10 7 m 1 = 1 911.3 A Bohr atom: E n = ke 2 Z 2 r n = E 0 Z 2 n 2 ; E 0 = ke 2 2 a 0 = m e ( ke 2 ) 2 h 2 = 13.6 eV ; K = m e v 2 2 ; U = ke 2 Z r hf = E i E f ; r n = r 0 n 2 ; r 0 = a 0 Z ; a 0 = h 2 m e ke 2 = 0.529 A ; L = m e vr = n h angular momentum de Broglie : λ = h p ; f = E h ; ω = 2 π f ; k = 2 π λ ; E = h ω ; p = h k ; E = p 2 2 m Wave packets: y ( x , t ) = a j cos( k j x ω j t ), or j y ( x , t ) = dk a ( k ) e i ( kx - ω ( k ) t ) ; Δ k Δ x ~ 1 ; Δ ω Δ t ~ 1 group and phase velocity : v g = d ω dk ; v p = ω k ; Heisenberg : Δ x Δ p ~ h ; Δ t Δ E ~ h Schrodinger equation: - h 2 2m 2 Ψ x 2 +U(x) Ψ (x,t) = i h Ψ t ; Ψ (x,t) = ψ (x)e -i E h t

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document