Unknown isotope Problem description:
The most common isotope of a single nucleus stripped of its electrons is accelerated through a potential difference of 1225V and fired horizontally into a B-field directed perpendicularly into the page (flat page lying on desk). The nucleus is observed to travel in a circular path of radius 9.973 cm. Afterward an alpha particle is accelerated through the same potential difference and fired in the same manner into the magnetic field and observed to travel in a circular path with radius 8.980 cm. What element is the unknown nucleus? (For this problem determine the masses of the nuclei--unknown and alpha--by summing the individual masses of the protons and neutrons. Also, include a diagram with your solution showing all aspects of the problem description.)
This is what I have so far:
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This is the point where I am confused as to how to find the mass and charge of the unknown particle, or where exactly I'm supposed to go from this point to solve the problem. (I have heard from other students that there is some guessing involved (from the periodic table))
Help would be appreciated.
Thank you.
 A: First part: From the formula for the radius, and the fact that magnetic field is the same in both cases, you get:
$$ B = \frac{m_1 v_1}{q_1 r_1} = \frac{m_2 v_2}{q_2 r_2} $$
Because you don't know the velocities, you want to get them from the potential difference. You also have
$$ v = \sqrt\frac{2q V}{m} $$
You put that back into the first equation, and isolate the two terms you don't know: $q_1$ and $m_1$:
$$ \frac{m_1}{q_1} = \frac{m_2}{q_2} \left(\frac{r_1}{r_2}\right)^2 $$
So, you have for your unknown particle $m/q = 2.4677$.
Note that you could also get there from your calculations, because you have calculated $B$. The above is simpler because by writing that $B$ and $V$ are constant, we use invariants to relate quantities relating to particles 1 and 2.

Second part, the search!
We don't want to check all elements one by one. To get an idea of where that element is, we take advantage of the fact that, for an isotope, mass is roughly equal to $A$ (atomic number). So, we want $A/Z \approx 2.4677$. Then, we use what we know about stable isotopes: they are situated mostly along a curve near (but not along) the $A=2Z$ axis. Check this diagram from Wikipedia:

The most stable isotopes (which include the most common isotope of each element) are the marroon ones. They are not randomly distributed, but follow a rough curve. Where does this curve intersect the line for $A/Z \approx 2.4677$? Well, you can check a few values here and there. In my case, I used Mathematica so you can see better the idea (the red line is $m/Z = 2.4677$):
Show[
 ListPlot[
  Table[ElementData[ElementData[i], "AtomicWeight"], {i, 1, 100}]],
 Plot[2.4677*x, {x, 0, 100}, PlotStyle -> Red]
 ]


The crossing is thus somewhere between 60 and 70. I have experimented myself and found that the answer is Gd (gadolinium, element 64). It is the element that best fits the bill: its most common isotope is 158Gd, has a mass of 157.9241039 and its $q/m$ ratio is thus: 2.4675. That's within the uncertainty of the measurements given.
