11. Determine the velocity (a vector quantity) of the scattered electron. As part of your response, sketch the situational diagram showing the path of the scattered electron, sketch a vector addition diagram consistent with the vector analysis method you are choosing, and state all necessary physics principles and formulas. Use the following information to answer this question. In an observation of the Compton effect a photon is incident on a free electron. The scattered photon is detected, as illustrated below. The scattered electron is not shown. Before After Stationary Path of free electron incident photon 82.80 A = 3.75 x 10 Path of scattered photon [pl = 1.67 x 10-N . S E = 5.02 x 1015 JPHYSICS DATA SHEET Constants Prefixes Used with SI Units Particles Acceleration Due to Gravity Exponential Charge Mass Near Earth....... lag| = 9.81 m/s2 Prefix Symbol Value Gravitational Constant ...... G=6.67 x 10-1 N-m2/kg Alpha Particle...... ......... + 2e 6.65 x 10 2 kg atto .... . a.. 10-18 Radius of Earth .... Te = 6.37 x 100 m femto ..... f ........ 10-15 Electron ... -le 9.11 x 10 kg Mass of Earth... ....... Me = 5.97 x 1024 kg pico............ P.............. 10-12 Proton ..................."... +le 1.67 x 10-27 kg Elementary Charge .......... e= 1.60 x 10-19 C nano....... n. ........... 10-9 Coulomb's Law Constant .. k= 8.99 x 109 N.m2/C2 micro ... /! ............ 10-6 Neutron..... 0 1.67 x 10-27 kg Electron Volt ....... ... 1 eV = 1.60 x 10-19 J milli...... m..........10-3 First-Generation Fermions Index of Refraction of Air. n = 1.00 centi...... C............. 10-2 Charge Mass Speed of Light in Vacuum. c = 3.00 x 10% m/s deci........ d.. .. 10-1 Electron .... -le ~0.511 MeV/c2 Planck's Constant ...... h = 6.63 x 10 34 J.s deka ......... da .... 101 h = 4.14 x 10 eV.s 102 Positron ..... +le ~0.511 MeV/c2 hecto ..... h. ...... Atomic Mass Unit ......... u = 1.66 x 10 2kg kilo ..... K......... 103 Electron neutrino, v........ 0 mv2 Rectangle = lw c2 = a2+62 FAt = mAv Ep = mgh E = 1 = 9 Triangle = Tab W = Fla|cose Ep = 5kx2 Circle = 172 Fe a b E = q |Fm| = 1 B sin A sin B sin C W = AE Circumference E = AV Fm| = qVB c2 = a2+ 62 -2ab cos C P = W Circle = 2ar Ad