# Radius of a lead nucleus and alpha particle

1. Nov 12, 2015

### Oribe Yasuna

1. The problem statement, all variables and given/known data
A lead nucleus contains 207 nucleons (82 protons and 125 neutrons) packed tightly against each other. A single nucleon (proton or neutron) has a radius of about 1 ✕ 10^−15 m.

(a) Calculate the approximate radius of the lead nucleus.

(b) Calculate the approximate radius of the alpha particle, which consists of 4 nucleons, 2 protons and 2 neutrons.

(c) What kinetic energy must alpha particles have in order to make contact with a lead nucleus?

2. Relevant equations
v_sphere = 4/3 pi r^3
r_nucleon = 1 * 10 ^ -15 m

3. The attempt at a solution
207 * 1 * 10 ^ -15 m
2.07 * 10 ^ -13 m

Beyond this, I don't have any idea what to do.
I considered using d = m/v, but stopped halfway because I was approaching a bizarre answer.

2. Nov 12, 2015

### Staff: Mentor

That would be the approximate length of a chain of 207 nucleons, spaced one radius apart each. That's not how a lead nucleus looks like. Where does that multiplication come from?

What are d, m and v?
Did you try conservation of energy?

3. Nov 12, 2015

### Oribe Yasuna

I figured the radius of a single proton/neutron is 1* 10 ^ -15 so the radius of the nucleus, which consists of 207 protons and neutrons, would be 207 multiplied by 1 * 10 ^ -15 m.

Since 207 doesn't have any units, dimensional analysis checked at least.

Density, mass and volume.
I thought of doing that because I could calculate the mass and volume, but of course density doesn't seem to have anything to do with the problem.

I didn't try the Law of Conservation of Energy (m_1 v_1 = m_2 v_2).
But I don't get how it would help? Doesn't it need at least 1 velocity vector? I only have volume, radius, and potentially mass.

Sorry for the late reply.

4. Nov 12, 2015

### haruspex

If I arrange 27 identical cubes of side 1cm into one large cube, how long is each side of the large cube?

5. Nov 12, 2015

### Oribe Yasuna

3cm.

6. Nov 12, 2015

### haruspex

Right. Not 27cm. So why would a cluster of 207 spheres each of radius 1* 10 ^ -15m have a radius of 207 times that?

7. Nov 12, 2015

### BvU

So if the number (the volume) increases by a factor of 3, the 'length' increases by a factor of 3.
Now a bulging stuffed bag with 207 oranges. what's the approximate raduis in units of one orange radius ?
 ah, Haru is back. Sorry for barging in ! Bedtime .

8. Nov 12, 2015

### Oribe Yasuna

They wouldn't.

So I took the cubed root of 207 (5.9154817) and multiplied it by 1*10^-15 m.
The answer I got was 5.915*10^-15 m and it was correct.

For part b)
I figured a cube of 4 smaller cubes would have sides of size 2. So I multiplied 5.9154817*10^-15 m by 2 to get 1.18*10^-14 m.
But the answer was wrong. Why? Is it just because I didn't put a fourth significant figure, maybe?

9. Nov 12, 2015

### Oribe Yasuna

I don't know why I multiplied by 2.

4^1/3 * 1*10^-15 m = 1.59*10^-15 m
which was the answer for part b).

10. Nov 12, 2015

### Oribe Yasuna

I don't understand how to get part c).

This is what I know:
r_lead = 5.91*10^-15 m
r_alpha = 1.59*10^-15 m
v = 4/3 pi r^3
K = 1/2 mv^2
1.6*10^-19 J = 1 eV
1 eV = 1*10^-6 MeV
m_proton = 1.6726219*10^-27
m_neutron = 1.6749*10^-24

I calculated v_sphere for the lead nucleus and alpha particle:
v_lead = 8.670795724*10^-43 m^3
v_alpha = 1.675516082*10^-44 m^3

I don't know where to go with this though.

11. Nov 12, 2015

### haruspex

Think about this: what makes it hard for the alpha particle to hit the nucleus?

12. Nov 12, 2015

### Oribe Yasuna

Its size?

13. Nov 12, 2015

### haruspex

I suppose that is one problem. But why should any particular energy be required? Why couldn't they just drift slowly together?

14. Nov 12, 2015

### Oribe Yasuna

Because protons repel each other and they're both made of protons, right?

Last edited: Nov 12, 2015
15. Nov 12, 2015

### haruspex

Right. So how does that lead to a specific energy requirement?

16. Nov 12, 2015

### Oribe Yasuna

You need enough kinetic energy to move against that repellent force.

So kinetic energy would be equal to the force that the protons repel each other with.
Would that be electrical potential energy?

17. Nov 12, 2015

### haruspex

Yes. Do you know any equations for that?

18. Nov 12, 2015

### Oribe Yasuna

U = 1/(4 pi E_knot) * (q_1 q_2) / r

1/ 4 pi E_knot = 9 * 10^9 Nm^2 / c^2
r = 5.92*10^-15 + 1.58*10^-15 = 7.50*10^-15

I don't know what q_1 and q_2 are though? Something to do with v_sphere?

19. Nov 12, 2015

### haruspex

They're the charges. Look up the charge of a proton and count the protons in each.

20. Nov 12, 2015

### Oribe Yasuna

q_1 = 82 * 1.6*10^-19 = 1.312*10^-17
q_2 = 2 * 1.6*10^-19 = 3.2*10^-19

U = 9*10^9 * (1.312*10^-17 * 3.2*10^-19) / 7.50*10^-15
U = 9*10^9 * 4.1984*10^-36 / 7.50*10^-15
U = 5.036144273*10^-12 J

1 eV = 1.6*10^-19 J

5.036144273*10^-12 / 1.6*10^-19 = 3.147590171*10^7 eV

1 MeV = 1*10^6 eV

3.14*10^7 / 1*10^6 = 31.48 MeV

It was correct!
Thanks a ton, I appreciate it.