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Radius of a lead nucleus and alpha particle

  1. Nov 12, 2015 #1
    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. jcsd
  3. Nov 12, 2015 #2

    mfb

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    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?
     
  4. Nov 12, 2015 #3
    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.
     
  5. Nov 12, 2015 #4

    haruspex

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    If I arrange 27 identical cubes of side 1cm into one large cube, how long is each side of the large cube?
     
  6. Nov 12, 2015 #5
    3cm.
     
  7. Nov 12, 2015 #6

    haruspex

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    Right. Not 27cm. So why would a cluster of 207 spheres each of radius 1* 10 ^ -15m have a radius of 207 times that?
     
  8. Nov 12, 2015 #7

    BvU

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    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:.
     
  9. Nov 12, 2015 #8
    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?
     
  10. Nov 12, 2015 #9
    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).
     
  11. Nov 12, 2015 #10
    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.
     
  12. Nov 12, 2015 #11

    haruspex

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    Think about this: what makes it hard for the alpha particle to hit the nucleus?
     
  13. Nov 12, 2015 #12
    Its size?
     
  14. Nov 12, 2015 #13

    haruspex

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    I suppose that is one problem. But why should any particular energy be required? Why couldn't they just drift slowly together?
     
  15. Nov 12, 2015 #14
    Because protons repel each other and they're both made of protons, right?
     
    Last edited: Nov 12, 2015
  16. Nov 12, 2015 #15

    haruspex

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    Right. So how does that lead to a specific energy requirement?
     
  17. Nov 12, 2015 #16
    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?
     
  18. Nov 12, 2015 #17

    haruspex

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    Yes. Do you know any equations for that?
     
  19. Nov 12, 2015 #18
    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?
     
  20. Nov 12, 2015 #19

    haruspex

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    They're the charges. Look up the charge of a proton and count the protons in each.
     
  21. Nov 12, 2015 #20
    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.
     
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