Radius of a lead nucleus and alpha particle

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SUMMARY

The discussion focuses on calculating the radius of a lead nucleus and an alpha particle, along with determining the kinetic energy required for alpha particles to contact a lead nucleus. The radius of the lead nucleus, containing 207 nucleons, is calculated to be approximately 5.91 × 10-15 m, while the radius of the alpha particle, consisting of 4 nucleons, is approximately 1.59 × 10-15 m. The kinetic energy needed for alpha particles to overcome the repulsive force from protons in the lead nucleus is found to be around 31.48 MeV, derived from the electrostatic potential energy equation.

PREREQUISITES
  • Understanding of nuclear physics concepts, including nucleons and their properties.
  • Familiarity with the formula for the volume of a sphere: v = 4/3 π r3.
  • Knowledge of electrostatic potential energy and the relevant equations.
  • Basic understanding of energy units, specifically electronvolts (eV) and mega-electronvolts (MeV).
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  • Learn about electrostatic forces and how they apply to charged particles in nuclear interactions.
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  • Investigate the relationship between energy units, focusing on conversions between joules, eV, and MeV.
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Students and professionals in physics, particularly those focusing on nuclear physics, particle physics, and anyone involved in calculations related to nuclear interactions and energy requirements in particle collisions.

Oribe Yasuna
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Homework Statement


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?

Homework Equations


v_sphere = 4/3 pi r^3
r_nucleon = 1 * 10 ^ -15 m

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.
 
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Oribe Yasuna said:
207 * 1 * 10 ^ -15 m
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?

Oribe Yasuna said:
I considered using d = m/v, but stopped halfway because I was approaching a bizarre answer.
What are d, m and v?
Did you try conservation of energy?
 
mfb said:
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?
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.

mfb said:
What are d, m and v?
Did you try conservation of energy?
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.
 
Oribe Yasuna said:
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.
If I arrange 27 identical cubes of side 1cm into one large cube, how long is each side of the large cube?
 
haruspex said:
If I arrange 27 identical cubes of side 1cm into one large cube, how long is each side of the large cube?
3cm.
 
Oribe Yasuna said:
3cm.
Right. Not 27cm. So why would a cluster of 207 spheres each of radius 1* 10 ^ -15m have a radius of 207 times that?
 
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 ?
[edit] ah, Haru is back. Sorry for barging in ! Bedtime :sleep:.
 
haruspex said:
Right. Not 27cm. So why would a cluster of 207 spheres each of radius 1* 10 ^ -15m have a radius of 207 times that?
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?
 
Oribe Yasuna said:
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?
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
Oribe Yasuna said:
(c) What kinetic energy must alpha particles have in order to make contact with a lead nucleus?
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
Oribe Yasuna said:
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.
Think about this: what makes it hard for the alpha particle to hit the nucleus?
 
  • #12
haruspex said:
Think about this: what makes it hard for the alpha particle to hit the nucleus?
Its size?
 
  • #13
Oribe Yasuna said:
Its size?
I suppose that is one problem. But why should any particular energy be required? Why couldn't they just drift slowly together?
 
  • #14
haruspex said:
I suppose that is one problem. But why should any particular energy be required? Why couldn't they just drift slowly together?
Because protons repel each other and they're both made of protons, right?
 
Last edited:
  • #15
Oribe Yasuna said:
Because protons repel each other and they're both made of protons, right?
Right. So how does that lead to a specific energy requirement?
 
  • #16
haruspex said:
Right. So how does that lead to a specific energy requirement?
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
Oribe Yasuna said:
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?
Yes. Do you know any equations for that?
 
  • #18
haruspex said:
Yes. Do you know any equations for that?
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
Oribe Yasuna said:
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?
They're the charges. Look up the charge of a proton and count the protons in each.
 
  • #20
haruspex said:
They're the charges. Look up the charge of a proton and count the protons in each.
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.
 
  • #21
Oribe Yasuna said:
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.
You are welcome.
 
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