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Homework Help: More Nuclear Equations

  1. Mar 21, 2008 #1
    Hi there. I'm hoping someone will be able to help with these equations. I'm studying 1st year chem through correspondence, and am struggling a little with some concepts without someone to talk them through with.

    These are the equations I'm given.

    i) [tex]^{28}_{14}Si \ + \ ^1_0n \ \rightarrow \ ^{25}_{12}Mg \ + \ X[/tex] (neutron bombardment?)

    ii) [tex]^{186}_{75}Re \ + \ ^1_1p \ \rightarrow \ X \ + \ ^0_{-1}e[/tex] (proton bombardment?)

    iii) [tex]^{208}_{83}Bi \ + \ ^0_{-1}e \ \rightarrow \ ^{207}_{81}Tl \ + \ X[/tex] (electron bombardment?)

    iv) [tex]^{14}_6C \ \rightarrow \ ^{13}_6C \ + \ X[/tex] (decay?)

    As far as i can figure, these are the answers. If someone could please point out if and where i might be going wrong I would appreciate it.

    i) [tex]X \ = \ ^4_2He[/tex]

    ii) [tex]X \ = \ ^{187}_{75}Re[/tex]

    iii) [tex]X \ = \ ^1_1p[/tex]

    iv) [tex]X \ = \ ^0_1n[/tex]

    I'm sure these seem pretty simple(not to me), but if you could verify my answers, it would really help me with some conceptual understanding.

    Thanks very very much in advance. :smile:
  2. jcsd
  3. Mar 21, 2008 #2


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    With [tex]{^A_Z}X[/tex], A is the atomic mass and Z is the atomic number or charge.

    The electron has charge -1, so Z=-1, but it does not contribute (much) to atomic mass since me ~ mp/1836

    In these equations, the sums of charges and masses on the left side must equal the sums on the right.

    In the first reaction (n, [itex]\alpha[/itex]), for A, one has 28 + 1 = 25 + A(X) or A(X) = 29 - 25 = 4, which would be the mass of a He atom (or alpha particle). Similarly, for Z, 14 + 0 = 12 + Z(X), or Z(X) = 14 - 12 = 2, so one has the correct answer for i.

    ii) A is correct, but check the charge balance.

    iii) correct

    iv) switch the subscript and superscript.

    The neutron is [tex]{^1_0}n[/tex], i.e. Z = 0, but A = 1.
  4. Mar 21, 2008 #3
    Hi Astronuc. thanks sooo much for this.

    Ok, so for ii, [tex]^{186}_{75}Re[/tex] gains a proton, increasing A by 1, and therefore Z by 1 as well, giving us [tex]^{187}_{76}[/tex]. In order for the charges to match, we would need [tex]X \ = \ ^{187}_{77}Ir[/tex].

    I must say it seemed unlikely that the answer would be a nuclide of the same element. :)

    I guess this may be an idiotic question, but i have to ask. [tex]^{186}_{75}Re[/tex] gains a proton, Z=187 A=76. Then, the nucleus gives off an electron, and this loss of 1 negative charge causes/allows a neutron to become a proton, increasing A by 1 and leaving Z alone. Does that make sense?

    With regards to the neutron in iv, thanks for that, got it :)

    Thanks so much for taking the time, and especially so fast.
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