130_notes.dvi

Sinceλ=h/p=h/Ein our fine units,

λ′−λ=

h

m

(1−cosθ).

We now apply the speed of light to make the units come out to be a length.

λ′−λ=

hc

mc^2

(1−cosθ)

These calculations can be fairly frustrating if you don’t decide which variables you want to keep and
which you need to eliminate from your equations. In this case we eliminatedφby using the energy
equation and computingp^2 e.

2.6.4 Rutherford’s Nuclear Size*.

If the positive charge in gold atoms were uniformly distributed over asphere or radius 5 Angstroms,
what is the maximumαparticle kinetic energy for which theαcan be scattered right back in the
direction from which it came?

To solve this, we need to compute the potential at the center of the charge distribution relative to
the potential at infinity (which we will say is zero). This tells us directlythe kinetic energy in eV
needed to plow right through the charge distribution.

The potential at the surface of the nucleus is 4 πǫ^10 ZeR where Z is the number of protons in the atom
and R is the nuclear radius. That’s the easy part. Now we need to integrate our way into the center.

V=

1

4 πǫ 0

Ze

R

−

∫^0

R

1

4 πǫ 0

r^3

R^3

Ze

r^2

dr

The r

3

R^3 gives the fraction of the nuclear charge inside a radiusr.

V=

1

4 πǫ 0

Ze

R

−

1

4 πǫ 0 R^3

∫^0

R

Zerdr

V=

1

4 πǫ 0

(

Ze

R

+

Ze

2 R

)

=

3 Ze

8 πǫ 0 R

So

V=

(3)(79)(1. 6 × 10 −^19 C)

8 π(8. 85 × 10 −^12 C^2 /Nm^2 )R

=

1. 7 × 10 −^7

R

Nm/C

The is then the kinetic energy in eV needed for a particle of charge +eto plow right through the
center of a spherical charge distribution. Theαparticle actually has charge +2eso we need to
multiply by 2. For a nuclear radius of 5 ̊Aor 5× 10 −^10 meters, we need about 680 eV to plow
through the nucleus. For the actual nuclear radius of about 5 Fermis or 5× 10 −^15 meters, we need
68 MeV to plow through.