Question

An alpha particle is a nucleus of helium. It has twice the charge and four times the mass of the proton. When they were very far away from each other, but headed toward directly each other, a proton and an alpha particle each had an initial speed of 8.2×10−3 c, where c is the speed of light. What is their distance of closest approach? There are two conserved quantities. Make use of both of them. (c = 3.00 × 108 m/s, k = 1/4πε 0 = 8.99 × 109 N · m2/C2, e = 1.60 × 10-19 C, m proton = 1.67 x 10-27 kg)

HW 4 Spring 2019-Chapter 23 Problem 23.20 15 of 16 Part A An alpha particle is a nucleus of helium. It has twice the charge and four times the mass of the proton. When they were very far away from each other, but headed toward directly each other, a proton and an alpha particle each had an initial speed of 8.2x10-3 c, where c is the speed of light. What is their distance of closest approach? There are two conserved quantities. Make use of both of them. (c 3.00 10 mis,k14me 08.99x 10 N mc160 1019 C,m proton-1.67 x 1027 k) 3.9 10-14m 34*10-14m 2.8 10-14m 44x10-14m Submit Request Answer Provide Feedback Next

Answer 1

Initial (P.E + K.E) = final (P.E + K.E) rightwardsdoublearrow KQq/r_1 + 1/2 mV_1^2 + 1/2 mV_1^2 = KQq/r^2 + 1/2 mV_2^2 + 1/2 mV_2^2 rightwardsdoublearrow 2 kq^2/r_1 + (m + M)V_1^2/2 = 2 kq^2/r^2 + (m + M)v_2^2/2 In above equation Q is replaced with equation. Now replacing M with 4m we get, rightwardsdoublearrow 2kq^2/r_1 + 5 mV_1^2/2 = 2kq^2/r_2 + 5 mV_2^2/2 rightwardsdoublearrow 4 kq^2/r_1 + 5 mV_1^2 = 4 kq^2/r^2 + 5 mV_2^2 Now, initial momentum = final momentum M.V_1 - m v_1 = mV_2 + V_2 rightwardsdoublearrow 3 mV_1 = 5 mV_2 Therefore v_2 = 0.6 v_1 Put v_2 = 0.6v_1 in above equation rightwardsdoublearrow 4 kq^2/r_1 + 5 mV_1^2 = 4 kq^2/r_2 + 5 m(0.6 v_1)^2 rightwardsdoublearrow 4 kq^2/r_1 + 3.2 mV_1^2 = 4 kq^2/r_2 rightwardsdoublearrow 1/r_1 + 0.8 mV_1^2/kq^2 = 1/r_2 r_1 almostequalto infinity and 1/r^2 almostequalto 0 \r\n Therefore r_2 = 1.25 kq^2/mV^2 = 1.25 times 8.99 times 10^9 times (1.60 times 10^-19/(1.67 time 10^-27) times (2.46 times 10^6)^2 = 2.84 times 10^-14 m Therefore option (c) is correct.