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Mass spectrometers are used to measure the masses of ions - AQA - A-Level Physics - Question 3 - 2020 - Paper 2

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Mass spectrometers are used to measure the masses of ions. Figure 3 shows one part of a mass spectrometer. A narrow beam consists of positive lithium ions travellin... show full transcript

Worked Solution & Example Answer:Mass spectrometers are used to measure the masses of ions - AQA - A-Level Physics - Question 3 - 2020 - Paper 2

Step 1

Draw two labelled arrows on Figure 3 to show the directions of the electric force F_E and the magnetic force F_M acting on A.

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Answer

  1. Draw an arrow pointing up to represent the electric force F_E acting on the positive lithium ion A due to the electric field E.
  2. Draw an arrow pointing downwards to represent the magnetic force F_M acting on the ion, which is in the direction according to the right-hand rule.

Step 2

Calculate E.

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Answer

Using the relationship for equilibrium of forces:

FE=FMF_E = F_M

Where:

  • Electric force, FE=qEF_E = qE
  • Magnetic force, FM=qvBF_M = qvB

Setting them equal gives: qE=qvBqE = qvB

Solving for E, we have:

E=vBE = vB

Substituting the known values:

  • v=1.5×105v = 1.5 × 10^5 m s1^{-1}
  • B=0.12B = 0.12 T

Thus, E=(1.5×105)×(0.12)=1.8×104V/mE = (1.5 × 10^5) × (0.12) = 1.8 × 10^4 \, \text{V/m}

Step 3

Show that the radius r of the circular path is given by r = mv/Bq.

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Answer

In a uniform magnetic field, the centripetal force needed for circular motion is provided by the magnetic force:

Fc=FMF_c = F_M

Where:

  • Centripetal force, Fc=mv2rF_c = \frac{mv^2}{r}
  • Magnetic force, FM=qvBF_M = qvB

Setting these equal gives: mv2r=qvB\frac{mv^2}{r} = qvB

Rearranging for r:

r=mvqBr = \frac{mv}{qB}

This shows that the radius r is directly proportional to mass m and inversely proportional to charge q and magnetic field B.

Step 4

Calculate the distance between the small aperture and the point where this ion strikes the detector surface.

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Answer

Using the formula for radius of the path:

r=mvqBr = \frac{mv}{qB}

Substituting known values:

  • v=1.5×105v = 1.5 \times 10^5 m s1^{-1}
  • B=0.12B = 0.12 T
  • Mass of 6Li+^{6}Li^+ ion = 1.0×10261.0 × 10^{-26} kg
  • Charge of 6Li+^{6}Li^+ ion, q=1.6×1019q = 1.6 × 10^{-19} C

Calculating r:

r=(1.0×1026)(1.5×105)(1.6×1019)(0.12)r = \frac{(1.0 \times 10^{-26}) (1.5 \times 10^5)}{(1.6 \times 10^{-19})(0.12)}

After calculations, find the distance from small aperture to detector surface.

Step 5

Determine the speed of the ion when it emerges through aperture P.

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Answer

Using energy conservation, the work done by the electric field equals the gain in kinetic energy:

qEd=12mv2qEd = \frac{1}{2} mv^2

Where:

  • dd = distance between electrodes = 3000 V
  • Charge of ion, q=1.2×1019q = 1.2 × 10^{-19} C
  • Mass of ion = 1.2×10261.2 × 10^{-26} kg

To find v, solve for speed:

v=2qEdmv = \sqrt{\frac{2qEd}{m}}

Substituting the values:

  • E=6000VE = 6000 \text{V}
  • d=3000Vd = 3000 \text{V}

Thus, v is determined after performing calculations.

Step 6

Discuss how the masses of the ions can be deduced from the measurement of these time intervals.

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Answer

The time taken for ions to travel from aperture P to the detector relates closely to their mass and charge due to their varying accelerations. The ions of different masses will have different speeds and consequently take different times.

The relationship can be described by:

t=lvt = \frac{l}{v}

Where t is the time interval, l is distance traveled, and v is speed derived from mass and charge. Therefore, by measuring these time intervals, we can deduce the mass of the ions as the time intervals will differ according to the equation relating mass, charge, and speed.

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