Question

Shown in the figure below is a re-creation of the Rutherford experiment. The "target" (green) nucleus is Platinum and has a charge of Q = 1.25e-17 Coulombs and a large mass. You fire a Lithium nucleus, with charge q = 4.806e-19 Coulombs and mass m = 1.162e-26 kg at a velocity of V1 = 1.87e+07 m/s from a location far away from the Platinum nucleus. At moment #2, the separation between the centers of the two nuclei is d2 = 6.693e-14 meters. We shall treat the Platinum nucleus as stationary throughout the process. Very far away, d1 00 V Moment #1 d₂ Moment #2 Vs=0 dz=? Moment #3 Calculate all the following: m/s The velocity of the small nucleus at moment #2 V2 = The distance at moment #3 d3 = m

Answer 1

ou Charge ou Platinum 1-25 X10+ c. =91 Charge lithium 4.806 XiDC. 792 At liftauce o clopet approaches Potential Euergy Kq192 d lithicem = 1.162 Xió 26 Kinetic Energy - £mv2 M = nass of okg d=qkq, 92 mv2 = 2X9x109 x 1.25X107 x 4.806x109 (1.162x1525) 1.87 xl3)? d =26.611 X1015 ΟΥ do 2.66 x1614 m. Aud vz = IN; = 1.87 x10m Ambe The lithium will come at same speed towarels target nuden uutil vzom at distance of cloſet appooch Initially hi V=O (Platinum V dla