Question

Rutherford fired a beam of alpha particles (helium nuclei) at a thin sheet of gold. An...

Rutherford fired a beam of alpha particles (helium nuclei) at a thin sheet of gold. An alpha particle was observed to be deflected by 90.0°; its speed was unchanged. The alpha particles used in the experiment had an initial speed of 1.6 ✕ 107 m/s and a mass of 6.7 ✕ 10−27 kg. Assume the alpha particle collided with a gold nucleus that was initially at rest. Find the speed of the nucleus after the collision.

Homework Answers

Answer #1

first, we consider the conservation of angular momentum in the x-direction.
m*v i = M*vfx

Where m is the mass of the alpha particle and vi is the initial velocity of the alpha particle. M is the mass of the gold nucleus and vfx is the velocity of the gold nucleus in the x-direction.
6.7 *10-27 *1.6*107 = 3.27 *10-25 * vfx

vfx = 3.27*105 m/s

The conservation of angular momentum in the y-direction.
0 = m*vy + M*vfy

0 = 6.7*10-27 *1.6*107 +  3.27 *10-25 * vfy

vfy = -3.27*105  m/s

This is directed along the negative y axis.
The net velocity of the gold nucleus
v = sqrt (vfx2 + vfy2 )
v = sqrt(2) * 3.27*105 m/s

v = 4.62*105 m/s

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