Solving Nuclear Equations: ^{28}_{14}Si to ^{14}_6C

[Phil Massie]

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) $$^{28}_{14}Si \ + \ ^1_0n \ \rightarrow \ ^{25}_{12}Mg \ + \ X$$ (neutron bombardment?)

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

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

iv) $$^{14}_6C \ \rightarrow \ ^{13}_6C \ + \ X$$ (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) $$X \ = \ ^4_2He$$

ii) $$X \ = \ ^{187}_{75}Re$$

iii) $$X \ = \ ^1_1p$$

iv) $$X \ = \ ^0_1n$$

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.

 

[Astronuc]

With $${^A_Z}X$$, 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 m_e ~ m_p/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, $\alpha$), 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 $${^1_0}n$$, i.e. Z = 0, but A = 1.

 

[Phil Massie]

Hi Astronuc. thanks sooo much for this.

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

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. $$^{186}_{75}Re$$ 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.