Worked examples · Sheet 3

Velocity selectors and mass spectrometers

Theory used: Velocity selector Mass spectrometer Electric force and magnetic force together Circular motion in a uniform magnetic field

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

In a velocity selector the two parallel plane plates are $d = 1.0 cm$ apart and the potential difference between them is $Δ V = 300 V$ ; the magnetic field, perpendicular to the electric field, equals $B = 0.15 T$ . Compute the speed of the particles that cross the selector without being deflected, and determine toward which plate the protons with speed $2.5 \cdot 1 0^{5} m/s$ are deflected, given that the electric force on a positive charge is directed toward the negative plate.

E = \frac{Δ V}{d}, q E = q v B ⟹ v = \frac{E}{B} = \frac{Δ V}{B d}

E v = \frac{300}{1.0 \cdot 1 0 ^{- 2}} V/m = 3.0 \cdot 1 0^{4} V/m = \frac{3.0 \cdot 1 0 ^{4}}{0.15} m/s = 2.0 \cdot 1 0^{5} m/s

v^{'} = 2.5 \cdot 1 0^{5} m/s > \frac{E}{B} ⟹ e v^{'} B > e E

Between two parallel plane plates the electric field is uniform, neglecting edge effects, and equals $\frac{Δ V}{d}$ . By the velocity selector the particles that pass without deflection are those for which the electric force and the magnetic force, which are opposite, have the same magnitude, that is those with $v = \frac{E}{B} = 2.0 \cdot 1 0^{5} m/s$ . The result depends neither on the charge nor on the mass.

A faster proton experiences a magnetic force greater than the electric force, which instead does not depend on the speed; therefore the magnetic force, opposite to the electric force, prevails, and the proton is deflected toward the positive plate.

Exercise 2

In a mass spectrometer $^{6} Li^{+}$ and $^{7} Li^{+}$ ions, of charge $e$ and masses $6.0 u$ and $7.0 u$ , start practically at rest, are accelerated by the potential difference $Δ V = 2.0 kV$ and enter a uniform magnetic field $B = 0.40 T$ perpendicular to their velocity. Compute the radii of the two trajectories and the distance between the points where the two isotopes hit the plate. Use $1 u = 1.66 \cdot 1 0^{- 27} kg$ .

e Δ V = \frac{1}{2} m v^{2}, r = \frac{m v}{e B} ⟹ r = \frac{1}{B} \frac{2 m Δ V}{e}, Δ x = 2 (r_{7} - r_{6})

r_{6} r_{7} Δ x = \frac{1}{0.40} \frac{2 \cdot 6.0 \cdot 1.66 \cdot 1 0 ^{- 27} \cdot 2.0 \cdot 1 0 ^{3}}{1.60 \cdot 1 0 ^{- 19}} m \approx 3.945 \cdot 1 0^{- 2} m = r_{6} \frac{7.0}{6.0} \approx 4.261 \cdot 1 0^{- 2} m = 2 \cdot (4.261 - 3.945) \cdot 1 0^{- 2} m \approx 6.3 \cdot 1 0^{- 3} m

The ions have the same charge and are accelerated by the same potential difference, so they have the same kinetic energy $e Δ V$ and different speeds. By the mass spectrometer the radius is proportional to the square root of the mass, therefore the ratio of the radii is $\frac{7}{6}$ , and it suffices to compute a single radius.

Each ion hits the plate after half a turn, at distance $2 r$ from the slit, so the distance between the two points of impact is twice the difference of the radii, about $6.3 mm$ . To increase the separation one can decrease the field, because $r$ is inversely proportional to $B$ , or increase the accelerating voltage.

Exercise 3

In a Bainbridge spectrometer ions of charge $e$ cross a velocity selector with electric field $E = 1.2 \cdot 1 0^{5} V/m$ and magnetic field $B_{1} = 0.30 T$ ; at the exit they enter a uniform magnetic field $B_{2} = 0.60 T$ , perpendicular to their velocity, and hit the plate $27.7 cm$ from the slit. Compute the mass of the ions, in kilograms and in atomic mass units, and the distance from the slit at which ions of the same charge with mass $22 u$ would arrive. Use $1 u = 1.66 \cdot 1 0^{- 27} kg$ .

The ions cross the selector (fields

E

and

B_{1}

) and, in the field

B_{2}

, describe semicircles of diameter

27.7 cm

for

20 u

and

30.4 cm

for

22 u

v = \frac{E}{B _{1}}, 2 r = \frac{2 m v}{e B _{2}} ⟹ m = \frac{e B _{1} B _{2} r}{E}, 2 r^{'} = \frac{2 m ^{'} E}{e B _{1} B _{2}}

v m m 2 r^{'} = \frac{1.2 \cdot 1 0 ^{5}}{0.30} m/s = 4.0 \cdot 1 0^{5} m/s = \frac{1.60 \cdot 1 0 ^{- 19} \cdot 0.60 \cdot 0.1385}{4.0 \cdot 1 0 ^{5}} kg \approx 3.32 \cdot 1 0^{- 26} kg \approx \frac{3.32 \cdot 1 0 ^{- 26}}{1.66 \cdot 1 0 ^{- 27}} u \approx 20.0 u = \frac{2 \cdot 22 \cdot 1.66 \cdot 1 0 ^{- 27} \cdot 1.2 \cdot 1 0 ^{5}}{1.60 \cdot 1 0 ^{- 19} \cdot 0.30 \cdot 0.60} m \approx 0.304 m = 30.4 cm

The velocity selector lets through only the ions with $v = \frac{E}{B _{1}}$ , so all the ions that enter the field $B_{2}$ have the same speed, $4.0 \cdot 1 0^{5} m/s$ . In the field $B_{2}$ , by circular motion in a uniform magnetic field, each ion describes half a circle of radius $\frac{m v}{e B _{2}}$ and hits the plate at distance $2 r$ ; the measured radius is $13.85 cm$ , from which the mass follows.

The mass is about $20.0 u$ , the value of the most abundant isotope of neon. For equal speed and charge the radius is proportional to the mass, so the ions of $22 u$ arrive at the distance $\frac{2 m ^{'} E}{e B _{1} B _{2}} \approx 30.4 cm$ from the slit, about $2.7 cm$ from the others. In this spectrometer the distance is proportional to the mass, while in the spectrometer with acceleration only it is proportional to the square root of the mass.