Physics 2524, Review Discussion Alpha Decay: Consider two charges, qt = 2e and q
ID: 1516805 • Letter: P
Question
Physics 2524, Review Discussion Alpha Decay: Consider two charges, qt = 2e and qy = 90c with ina.es ml = 4m and m2 = 234m separated by 8 fm (I fin = fennometer = m), where e = 1.602 x 10-19 C is the fundamental unit of charge and m = 1.67 x 10-. kg is the mass of a proton. This is a simple model of radioactive decay of a uranium nucleus into a thorium nucleus (90 protons and 144 neutron, and an alpha particle (2 protons and 2 neutrons). We are ignoring the influence of the electrons in this model. (a) At what position, relative to particle 1 is the magnitudes of the electric field due to the individual particles equal? Keeping the calculation in terms of c, m, and fm keeps the calculation easier. (b) Find the electric periled due to both charges along the line that bisects the line connecting the charges. (Let the charges lie aim% the x-axis on equal sides of the y-axis and find the field along the y-axis.) Leave answer in tern. of k, c, and in (clinometers. (c) Find the electrostatic force that particle 2 exerts on particle I. Find the magnitude of the acceleration of particle 1. Find both answers in SI units. (d) Find the elicit potential energy of the system in Joules. (e) What is the velocity (in m/s) of the alpha particle (particle 1) when it is very far from the nucleus? Assume the system is initially at rest and that the thorium nucleus (particle 2) stays at rest due to its much larger mass. Anything wrong v4th number?Explanation / Answer
q1 = 2e = 2*1.6*10^-19 = 3.2*10^-19 C
q2 = 90e = 90*1.6*10^-19 = 1.44*10^-17 C
d= 8fm = 8*10^-15 m
m1=4m= 4*1.67*10^-27 kg = 6.68*10^-27 kg
m2=234m= 234*1.67*10^-27 kg = 3.91*10^-25 kg
a)
Assume at point P , distance x from q1 , Enet = 0
Enet = E1 + E2 = k(q1/r1^2 + q2/r2^2)
Fnet = k(2e/x^2 + 90e/(d-x)^2]
0 = (9*10^9)[(3.2*10^-19)/(x)^2 + (1.44*10^-17)/( 8*10^-15 - x)^2] => x= 1.04*10^-15 m = 1.04fm
b)
Vnet = V1+V2 = kq1/r1+kq2/r2
since r1=r2=r=d/2
Vnet = V1+V2 = kq1/r+kq2/r = k/r(q1+q2) = (2k/d)(q1+q2) = (2k/d)(2e+90e) = (2k/d)(92e) = [(2*9*10^9)/(8*10^-15)]*(92*1.602*10^-19) = 3.3*10^7 V
c)
F12 = kq1q2/d^2 = (k*2e*90e)/d^2 = (180ke^2)/d^2 = (180*9*10^9*1.602*10^-19)/(8*10^-15)^2 = 4.06*10^21 N
a12 = F12/m1 = (4.06*10^21N)/(6.68*10^-27kg) = 6.08*10^47 m/s^2
d) EPE = kq1q2/d = (k*2e*90e)/d = (180ke^2)/d = (180*9*10^9*1.602*10^-19)/(8*10^-15) = 3.24*10^7 J
e) At very for from the nucleus, d= infinity
EPE = kq1q2/d = 0 J
Thus KEi = 0 J and hence vi= 0 m/s