A mass spectrometer sorts atoms by mass in two steps: a voltage gives every ion the same energy, then a magnetic field bends each into a circle, as in Charged Particles in Circles. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
$$qV = \tfrac{1}{2}mv^2, \qquad qvB = \frac{mv^2}{r} \ \Rightarrow\ r = \frac{mv}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}.$$ So $r \propto \sqrt{m}$ for ions sped up by the same voltage, and the time for a half circle is $\pi m/(qB)$.
Worked example: Neon ions
Singly charged neon-20 ions ($20\ \text{u}$) are sped up through $1000\ \text{V}$ and enter a $0.20\ \text{T}$ field. Find (1) their speed and (2) the radius.
- (1) $m = 20(1.66 \times 10^{-27}) = 3.32 \times 10^{-26}\ \text{kg}$; $v = \sqrt{2(1.6 \times 10^{-19})(1000)/m} = 9.82 \times 10^{4}\ \text{m/s}$.
- (2) $r = \dfrac{mv}{qB} = 10.2\ \text{cm}$.
Answer: $9.82 \times 10^4\ \text{m/s}$, $10.2\ \text{cm}$
Common mistake
Saying a heavier ion's radius grows in proportion to its mass. The heavier ion also leaves the voltage slower, so $r$ grows only as $\sqrt{m}$. Check: substitute $v$ before comparing.