# More Nuclear Equations

1. Mar 21, 2008

### 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.

2. Mar 21, 2008

### Astronuc

Staff Emeritus
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 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, $\alpha$), 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.

3. Mar 21, 2008

### 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.