Worked exercise — mass spectrometer for cyanide

Problem

A chemist analyses cyanide (CN$^-$) in a mass spectrometer: the carbon (C) and nitrogen (N) atoms are ionised, accelerated up to a kinetic energy $K = 10^{-4}\;\text{J}$ each, and then sent into a region with $|\vec{B}| = 1\;\text{T}$ perpendicular to the velocity. What are the radii of the circles traced by the two ions? The masses are $m_\text{C} = 12\,u$ and $m_\text{N} = 14\,u$, with $1\,u \approx 1{,}66\cdot 10^{-27}\;\text{kg}$.

Solution

Radius as a function of kinetic energy. Using $r = mv/(qB)$. From $K = \tfrac{1}{2}mv^2$ we get $v = \sqrt{2K/m}$, hence $mv = m\sqrt{2K/m} = \sqrt{2Km}$. Then: $r = \frac{\sqrt{2\,K\,m}}{q\,B}$

With $q = e = 1{,}6\cdot 10^{-19}\;\text{C}$ and $B = 1\;\text{T}$: $r_\text{C} = \frac{\sqrt{2\cdot 10^{-4}\cdot 12\cdot 1{,}66\cdot 10^{-27}}}{1{,}6\cdot 10^{-19}} = \frac{\sqrt{3{,}98\cdot 10^{-30}}}{1{,}6\cdot 10^{-19}} \approx \ev{1{,}25\cdot 10^4\;\text{m} = 12{,}5\;\text{km}}$

For nitrogen, since $r \propto \sqrt{m}$ at equal $K$, $q$, $B$: $r_\text{N} = \sqrt{\frac{14}{12}}\cdot r_\text{C} \approx \ev{1{,}35\cdot 10^4\;\text{m} = 13{,}5\;\text{km}}$

The difference in radii (about $1\;\text{km}$ out of $\sim 13\;\text{km}$) is proportional to $\sqrt{m_\text{N}/m_\text{C}}$. The spectrometer separates C and N with a relative resolution of $\sim 8\%$, sufficient to identify both.

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Topics: Magnetismo Concepts: Forza di Lorentz · Moto circolare uniforme