Download pdf file - High Point University

Question (2g30001)
Acceleration of an alpha particle by charged plates
An alpha particle is accelerated by two closely spaced, oppositely charged plates, as shown below.
Figure 1: An alpha particle moving between oppositely charged plates.
The alpha particle has a speed of 1.0 × 106 m/s when it enters a slit in the positively charged plate. After
traveling for 1 mm, it passes through a slit in the negatively charged plate. If the magnitude of the charge of
each plate is 10 µC, and if each plate has an area of 10 cm2 , what will be the speed of the alpha particle when
it reaches the negatively charged plate? (Note: the plate separation is small compared to the dimensions of
the plates.)
Solution
Dene a coordinate system with the +x axis in the direction of the velocity of the alpha particle. Dene two
points i and f along the path from the positively charged plate to the negatively charged plate.
Figure 2: Charged plates with coordinate system, initial and nal points, and the constant electric eld.
Let's begin by converting all units to m, kg, s, C, and combinations thereof. The charge on each plate has
a magnitude 10 × 10−6 C. The area of each plate is (10 cm2 )(1 m2 /1002 cm2 ) = 1 × 10−3 m2 . The plate
separation is 1 mm, or 0.001 m.
Choose the Fundamental Principle that can be used to solve this problemConservation of Energy. The
system of plates and particle are a closed system. Thus, the change in the total energy of the system is zero.
∆E
∆U + ∆K
∆K
=
0
= 0
= −∆U
Because the force on the alpha particle by the electric eld is in the same direction as the velocity of the
particle, then the particle will speed up as it travels toward the negatively charged plate. As a result, its
kinetic energy increases. Conservation of Energy tells us that this must be accompanied by a decrease in
potential energy. It is the loss of electric potential energy that results in a gain in kinetic energy. Conceptually,
we think of this as
|∆K|gained
= |∆U |lost
If we calculate the initial kinetic energy and the change in electric potential energy, then the nal kinetic
energy and nal speed can be determined.
(Kf − Ki )
Kf
= −∆U
= Ki − ∆U
The alpha particle is a helium nucleus with 2 neutrons and 2 protons; it's atomic mass is 4 g/mol which is
6.64 × 10−27 kg. Its initial kinetic energy is
Ki
1
mv 2
2 i
1
(6.64 × 10−27 kg)(1 × 106 m/s)2
=
2
= 3.32 × 10−15 J
=
The change in potential energy of the alpha particle (that has a charge of 2qproton ) from point i to point f
is:
∆U
= q∆V
= q(−Ex ∆x)
|Q|/A
= q −
∆x
0
= −(2)(1.6 × 10−19 C)
= −3.616 × 10−13 J
Thus, the nal kinetic energy is
10 × 10−6 C/1 × 10−3 m2
0
(0.001 m)
Kf
= Ki − ∆U
=
3.32 × 10−15 J − −3.616 × 10−13 J
=
3.32 × 10−15 J + 3.616 × 10−13 J
=
3.649 × 10−13 J
From this, calculate the nal speed, using m = 6.64 × 10−27 kg.
Kf
vf
1
mvf2
and solving for the nal speed gives:
2
r
2Kf
=
m
r
2Kf
=
m
= 1.05 × 107 m/s
=