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Question (1220002)
Impulse on an alpha particle by a gold nucleus (Rutherford scattering)
An alpha particle has a speed of
deects at an angle
52.9
◦
6.94 × 106
m/s in the +x direction when it collides with a gold nucleus. and
above the +x axis. Its speed after the collision nearly the same as its speed before
the collision because the gold nucleus is so massive compared to the alpha particle. What is the impulse on
the alpha particle due to the collision with the gold nucleus?
Solution
(a)
Begin by drawing a picture of the situation, showing the momentum of the alpha particle before and
after the collision.
Figure 1: The momentum of the alpha particle before and after it collides with a gold nucleus.
Before doing any calculations, we will need to know the initial velocity and nal velocity of the alpha particle
so that we can calculate its initial and nal momentum. The initial velocity is
~vi
= h6.94 × 106 , 0, 0i m/s
We are given the speed and angle of the nal velocity, so use direction cosines to get the components.
vf,x = |~v | cos(θx )
and
vf,y = |~v | cos(θy )
Sketch the nal velocity vector and the given angle.
The angle
θx
that the vector makes with the +x axis is
vf,x
52.9◦ ,
and the angle
= |~v | cos(θx )
=
( 6.94 × 106 m/s) cos(52.9◦ )
=
4.19 × 106 m/s
θy
is
90◦ − 52.9◦ = 37.1◦ .
Thus,
Figure 2: The nal velocity of the alpha particle after colliding with the gold nucleus.
and
vf,y
= |~v | cos(θy )
=
( 6.94 × 106 m/s) cos(37.1◦ )
=
5.53 × 106 m/s
An alpha particle is a helium nucleus. The mass of the electrons are negligible compared to the mass of the
protons and neutrons in the nucleus. To nd the mass of helium in kg, nd its mass in grams/mole on a
periodic table and convert this to kg using Avogadro's number.
m
He atom
=
=
4.003 g
mol
1 kg
1000 g
1 mol
6.02 × 1023 atoms
6.65 × 10−27 kg
Therefore, the initial momentum of the alpha particle is:
p~i
= m~vi
=
(6.65 × 10−27 kg)(h6.94 × 106 , 0, 0i m/s)
= h4.61 × 10−20 , 0, 0i kg m/s
The nal momentum of the alpha particle is
p~f
= m~vf
=
(6.65 × 10−27 kg)(h4.19 × 106 , 5.53 × 106 , 0i m/s)
= h2.79 × 10−20 , 3.68 × 10−20 , 0i kg m/s
According to the
momentum principle
∆~
p = F~net ∆t
The quantity
F~net ∆t
is called
impulse.
change in momentum of the puck.
Thus, one way to know the impulse is by simply measuring the
F~net ∆t
= ∆~
p
= p~f − p~i
= h2.79 × 10−20 , 3.68 × 10−20 , 0i kg m/s − h4.61 × 10−20 , 0, 0i kg m/s
= h−1.82 × 10−20 , 3.68 × 10−20 , 0i kg m/s
= h−1.82 × 10−20 , 3.68 × 10−20 , 0i N s
The units are expressed in
Ns
to remind us that this quantity represents force
×
time interval.
Now, check your answer. Sketch the change in momentum vector by placing the initial and nal momentum
vectors tail to tail and drawing the change in momentum from the head of the initial momentum vector to
the head of the nal momentum vector. The net force on the alpha particle is in the same direction as the
change in momentum of the alpha particle.
Figure 3: The impulse on the alpha particle by the gold nucleus.
In this case, the only force acting on the positively charged alpha particle is the repulsion of the positively
charged gold nucleus. The force on the alpha particle by the gold nucleus at the point of closest approach
(i.e. contact, since the nuclei never actually touch") is shown below.
Figure 4: The repulsive force on the alpha particle by the gold nucleus.