Question

A double charged unknown ion is shot into a 0.845 T magnetic field between two parallel plates which are 4.00 mm apart. When a potential difference of 56.8 kV is applied across the plates, the particle passes undeflected through the fields. When the voltage is turned off, the particle has a radius of curvature of 4.15 m.

Identify the ion

Answer 1

Solution: When an ion moves in a magnetic field it is affected by the Lorentz force

(1) $\vec F_L=q\vec v \times\vec B$ which in the first case is balanced by the Coulomb force exerted on the ion in the field of the capacitor

(2) $\vec F_c=q\cdot \vec E$

All vectors in this problem are directed so (perpendicular to each other) that only their magnitudes can be used. We have

(3) $F_c=F_L\\ qE=qvB$ and we are able to determine the speed

(4) $v=\frac{E}{B}$

When the voltage is turned off an ion moves about a ring with  Larmor or cyclotron radius [1]

(5) $r=\frac{mv}{qB}$

Thus we can determine the ratio $\frac {m}{q}$ of ion

(6) $\frac {m}{q}=\frac {Br}{v}={substitude (4)}=\frac {r B^2}{E}$

Given that the electric field in a capacitor can be expressed as $E=V/d$ we have

(7) $\frac {m}{q}=\frac {rd B^2}{V}=\frac{4.15m\cdot 4.00 \cdot 10^{-3} m\cdot (0.845 T)^2}{56.8\cdot 10^3 V}=2.09\cdot 10^{-7} \frac {kg}{C}$

Since it is known that an ion has a double charge $q=2\cdot 1.6\cdot 10^{-19}C=3.2\cdot 10^{-19}C$ we can get its mass

(8) $m=q\cdot (\frac{m}{q})= 3.2\cdot 10^{-19}C \cdot 2.09\cdot 10^{-7} \frac{kg}{C}=6.69 \cdot 10^{-26} kg$

and the atomic mass [2]

(9) $m=\frac{6.69\cdot 10^{-26}}{1.66\cdot 10^{-27}}u=40.3u$

From the periodic table the closest double ion with this mass is the calcium ion - $Ca^{+2}$

Answer: This is $Ca^{+2}$ ion.

[1] https://en.wikipedia.org/wiki/Gyroradius

[2] https://en.wikipedia.org/wiki/Dalton_(unit)