Q1. Here is a diagram of a mass spectrometer

(a) The magnetic field strength in the velocity selector is 0.14T and the electric field strength is 20 kVm^–1.

(i) Define the unit for magnetic flux density, the tesla.

The Tesla is the unit of flux density. It is the flux density, acting perpendicular to a wire,that produces a force of 1N on a metre of wire that is carrying a current of 1A .

1T = 1 N m^-1 A^-1 put into words

(2 marks)

(ii) Show that the velocity selected is independent of the charge on an ion.

In the velocity selector the magnetic field and elecric field act in opposition cancelling out the effect of each other for those charges travelling at the selected velocity. Therefore ions travelling at that velocity carry on in a straight line, the others are deflected and don't get through the collimator.

F_E = QV/d

F_M = BQv

F_E = F_M therefore QV/d = BQv

Q cancels so V/d = Bv and

v = V/Bd = E/B

v is therefore not dependent on Q

(2 marks)

(iii) Show that the velocity selected is about 140 kms^–1

v = E/B

v = 20 x 10³/0.14

= 1.43 x 10⁵

= 140 km s⁻¹ (to 2 sig figs) QED

(1 mark)

(b) A sample of nickel is analysed in the spectrometer. The two most abundant isotopes of nickel are:

and

Each ion carries a single charge of +1.6 × 10^–19 C

The Ni-58 ion strikes the photographic plate 0.28 m from the point P at which the ion beam enters the ion separator.

Calculate:

(i) the magnetic flux density of the field in the ion separator;

Magnetic force acts as a centripetal force:

BvQ = mv²/r

so the magnetic flux density B = ^mV/_rQ

mass of ion:

58 x 1.661 × 10^–27 kg = 9.64 x 10^-26 kg

B = (9.64 x 10^-26 x 1.43 x 10⁵) / (0.14 x 1.6 × 10^–19) = 0.615

B = 0.62 T

(3 marks)

(ii) the separation of the positions where the two isotopes hit the photographic plate.

If you look at the diagram you will see that the ions travel in a semicircle to the plate. We therefore have to work out the diameter of the paths and subtract them from each other to work out the distance they are apart.

We are given the diameter of the path of the first ion - 0.28 m

Radius r of the path is proportional to the mass m of ion

r ∝ m

∴ r₁/r₂ = m₁/m₂

r₂ = m₂/m₁ x r₁

new radius = ⁶⁰/₅₈ × 0.140 = 0.145 m

new diameter = 0.290 m

distance apart = 0.29 - 0.28 = 0.010 m

(2 marks)

(Total 10 marks)