In the early part of the 20th century, Sir Joseph Thomson (discoverer of the ele
ID: 1585736 • Letter: I
Question
In the early part of the 20th century, Sir Joseph Thomson (discoverer of the electron) proposed a plum puding model of the aton. He beleved that he poserve aarge tha on] a prea at ite ike plums Rutherford fired a beam of alpha partidles (helium nudles) at a thin metal sheet. Because alpha partides are positive, the plum pudding model predicted that they should only be slightly deflected by the positive pudding. Instead, Rutherford found that the alpha particdes were greatly deflected, and some even reversed direction completely. Rutherford was surprised by his results and said that it was ike firing a bullet at a tissue paper and seeing it bounce back. Rutherford concluded that the positive charge of the atom was not spread out ike pudding, but rather was concentrated in the center or nucleus of the atom. The alpha partides used in the experiment had an initial speed of 2 x 10 m/s and a mass of 6.7 x 10-27 kg. (a) Assuming the nucleus is initially at rest, the collsion is head-on and elastic, and the nudeus is much more massive than the alpha particle, find the final speed of the alpha particle after it collides with the nucleus 20000000m/s (b) Rutherford used gold in his experiment Check the assumption that the nudeus is much more massive than the alpha particle by finding the speed of the madeus after the colision 579400 Your response dners from the correct answer by more than 10%. Double check your calculations m/sExplanation / Answer
ELASTIC COLLISION
m1(alpha) = 6.7*10^-27 kg m2(nucleus) = 79*1.67*10^-27 kg
speeds before collision
v1i = 2*10^7 m/s v2i = 0 m/s
speeds after collision
v1f = ? v2f = ?
initial momentum before collision
Pi = m1*v1i + m2*v2i
after collision final momentum
Pf = m1*v1f + m2*v2f
from momentum conservation
total momentum is conserved
Pf = Pi
m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)
from energy conservation
total kinetic energy before collision = total kinetic energy after collision
KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2
KEf = 0.5*m1*v1f^2 + 0.5*m2*v2f^2
KEi = KEf
0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)
solving 1&2
we get
v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)
v1f = ( (6.7*10^-27-79*1.67*10^-27)*2*10^7 + (2*79*1.67*10^-27*0))/(6.7*10^-27+79*1.67*10^-27)
V1f = 2*10^7 m/s
part(b)
v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)
v2f = (2*6.7*10^-27*2*10^7))/((6.7*10^-27)+(79*1.67*10^-27))
v2f = 0.2*10^7 m/s