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Homework Help: How do you determine the charge of nuclei?

  1. May 10, 2007 #1
    I have a few questions about nuclear physics:

    How do you determine the charge of nuclei? Isn't it always positive?

    Please tell me what equations/concepts I need to know for these objectives:

    1. Use conservation of mass number and charge to complete nuclear reactions.

    2. Determine the mass number and charge of a nucleus after it has undergone specified decay processes.

  2. jcsd
  3. May 10, 2007 #2
    1 u is equal to approx. 931 MeV right?

    How come a table says that: 1u = 931 MeV/c^2?? Is this table wrong?
  4. May 10, 2007 #3
    Also, the mass number of an element is not equal to its atomic mass right??
    The mass number is the # of nucleons while the atomic mass is the total mass of the atom??

    To calculate the mass defect (i.e. Sum of individual nucleons - atomic mass) all you do is add up the masses of the protons and neutrons right? You don't need to worry about the mass of the electron since it is so small??
  5. May 10, 2007 #4
    Also, if the problem gives you an element with its mass number, how do you determine its atomic number of that element?? Do you need to look at the periodic table of elements?
  6. May 10, 2007 #5
    Yes it will be always positive, but the specific charge can be deduced from the number electrons the element has.
  7. May 10, 2007 #6


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    The nucleus of an atom consists of protons (+ charge) and neutrons (0 charge). The nucleus is surrounded by electrons (- charge). The charge of the proton and electron have the same magnitude, but different polarity or sign. The atomic number, Z, gives the number of protons or positive charges in the nucleus, and this is usually balanced by the same number of electrons around the nucleus to preserve charge neutrality.

    The mass number, A, gives the number of nucleons in the nucleus, and it is closely related to the atomic mass, since the proton has a mass of ~938.272 MeV/c2, or (1.6726 × 10−27 kg), 1.007 276 467 amu, while the neutron has mass 939.573 MeV/c2, 1.6749 × 10−27 kg), or1.008665 amu. Now notice that both the neutron and proton have mass > 1 amu. When combining into nuclei, some mass-energy is lost from the combination of nucleons, which we called binding energy.

    The use of MeV/c2 reflects Einstein's famous E = mc2, or m = E/c2 - a statement of the equivalence of mass and energy.

    The binding energy is determined by the differences in masses of the reactant particle/nucleus and the products of the reaction. One can use atomic mass as a reasonable approximation, since the number of electrons is preserved and the masses cancel out. However, if one is rigourous, one must account for the differences binding (ionization) energies of the electrons, but that is a relatively small number. Note that an electron mass is about 0.511 MeV/c2, and the ionization energies of the outermost electrons are of the order of a few eV, while that energy increases with Z for the inner most electrons, as is evident from characteristic X-ray energies.

    These might be helpful to get one started -



  8. May 10, 2007 #7
    Astronuc..ur replies r just superb and complete...u eat all those questions..which r related to nuke physics....

    My question is---
    Why r most lines of spectum (H atom) r so congested in I.R region...i.e. why they all end up in I.R. ?
  9. May 10, 2007 #8


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    Thank you for the kind words, krateesh. :smile:

    Well, let's take a look at the hydrogen spectrum -



    By itself, the hydrogen atom has 1 proton and 1 electron, and the electron configuration is 1s1. If the electron gets excited [by atomic collision or strong EM interaction] and falls back into its ground state, n = 1, this yields UV photon (Lyman series or lines). If the electron falls into the orbital (energy state) corresponding to n=2, the emitted photon is in the visible range (Balmer lines). For n=3, we have the Paschen series, which is infrared (IR). There are however, probably more energy transitions in the IR region than the limited number of UV and visible transitions.

    Now, under earth type conditions (as opposed to a stellar plasma), hydrogen tends to form a diatomic molecule, and IIRC most molecules tend to emit/absorb in IR or microwave. The principle of the microwave oven is the absorption (excitation) of specific microwave frequencies by the water 'molecule'.
  10. May 10, 2007 #9
    Yeah Astronuc, your replies are wonderful and they answer my questions perfectly. I have some other questions about energy levels. So of the Energy levels are:

    n1=-62eV, n2=-15eV, n3=-7, n4=-4, n5=-3

    The problem is: What could happen if this atom, while in an undetermined energy state, were bombarded with a photon of energy 10eV?

    The answer is: Since no two energy levels in this atom are separated by 10eV, the atom could not absorb a 10eV photon, and a result, nothing would happen. This atom would be transparent to light of energy 10eV

    My question: So the atom could only absorb photons with energies that match exactly to the energy difference of the levels? If any extra energy is absorbed then nothing would happen to the atom and the photon wouldn't be absorbed? Does this have to do with the equation: [tex] KE_{max}=E_{photon}-\phi[/tex]??

    --Does it also related to this sentence?: Why do atoms emit (or absorb) radiation only at certain discrete wavelengths?

    -->So what they mean by discrete wavelength is that the E difference of each level = hc/lambda and therefore only photons with specific wavelengths would be able to be absorbed by electrons?
  11. May 10, 2007 #10
    Oh yeah, how come the table says that 1u = 931 MeV/c^2?? What's up with the "extra" c^2?? I don't get why it is there and if it will affect my answer much if I don't convert something. Can you leave the c^2 out??

    I'm using the energy of 1u, which is approx. 931 eV to calculate the binding energy of other nucleons.

    [tex] \Delta m_{defect} * \frac{931 eV}{1u}[/tex]

    Does anyone know how to get 931 eV \ 1u , without the c^2??

    I'm just plugging in 1u= 1.66 x 10^-27kg to the equation: E=mc^2 to determine the energy of 1 atomic mass unit.
    Last edited: May 10, 2007
  12. May 11, 2007 #11


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    From post #6,

    The use of MeV/c2, reflects Einstein's famous E = mc2, or m = E/c2.

    MeV is a unit of energy. And be careful, 1 u has an energy equivalent of 931.478 MeV.

    If 1 u = 1.66 x 10-27 kg, then using E = mc2, one has

    E (1 u) = 1.66 x 10-27 kg * (2.99793 x 108 m/s)2 = 1.492 x 10-10 kg-m2/s2. But then use the definition that 1 N = 1 kg-m/s2 and 1 N-m = 1 J,

    so E (1 u) = 1.492 x 10-10 J.

    Then we use the equivalence between J and MeV: 1 MeV = 1.602 x 10-13 J,
    then E (1 u) = (1.492 x 10-10 J)/(1.602 x 10-13 J/MeV) ~931.3 MeV (with some round-off error).
    Last edited: May 12, 2007
  13. May 11, 2007 #12


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    With n3=-7, n4=-4, n5=-3 eV, then a 10 eV photon would cause the electron to move out and there would be energy left over - as in the photoelectric effect.

    Certainly in emission, the light produced is contrained by the difference in energy leves of an atom, or by the energy of the transition. In absorption, the same thing in reverse - light that matches a transition will cause that transition, but with less energy, the photon is simply scattered.

    One my wish to browse this page - http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html

    and browse the site for other details.
  14. May 11, 2007 #13
    Thanks for the replies. I really appreciate it.
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