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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 travelli... 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 upwards from the positive lithium ion A to represent the electric force FEF_E acting in the direction of the electric field.
  2. Draw an arrow pointing perpendicular to this, towards the left (into the page), to represent the magnetic force FMF_M acting on the positively charged lithium ions due to the magnetic field.

Step 2

Calculate $E$.

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Answer

Using the equation for the balance of forces, where the electric force equals the magnetic force:

FE=FMF_E = F_M

This gives:

qE=qvBqE = qvB

Simplifying, we can find EE:

E=vBE = vB

Substituting the known values:

E=(1.5imes105extms1)(0.12extT)=1.8imes104extVm1E = (1.5 imes 10^5 ext{ ms}^{-1})(0.12 ext{ T}) = 1.8 imes 10^4 ext{ V m}^{-1}.

Step 3

Show that the radius $r$ of the circular path is given by $r = \frac{mv}{Bq}$.

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Answer

The centripetal force required to keep an ion moving in a circular path in the magnetic field is provided by the magnetic force:

  1. We set the centripetal force equal to the magnetic force:

mv2r=Bqv\frac{mv^2}{r} = Bqv

  1. Re-arranging this equation gives:

r=mvBqr = \frac{mv}{Bq}.

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 we derived for the radius of the path:

  1. Substitute in the known values. The mass of the ion m=1.0imes1026extkgm = 1.0 imes 10^{-26} ext{ kg}, v=1.5imes105extms1v = 1.5 imes 10^5 ext{ ms}^{-1}, B=0.12extTB = 0.12 ext{ T}, and q=e=1.6imes1019extCq = e = 1.6 imes 10^{-19} ext{ C} (charge of Lithium ion).

Calculate rr:

r=mvBq=(1.0imes1026extkg)(1.5imes105extms1)(0.12extT)(1.6imes1019extC)r = \frac{mv}{Bq} = \frac{(1.0 imes 10^{-26} ext{ kg})(1.5 imes 10^5 ext{ ms}^{-1})}{(0.12 ext{ T})(1.6 imes 10^{-19} ext{ C})}

Calculating yields: r=0.0781extm=0.0781extm=7.81extcmr = 0.0781 ext{ m} = 0.0781 ext{ m} = 7.81 ext{ cm}.

Step 5

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

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Answer

Using the principle of energy conservation, the kinetic energy (KE) gained by the ion is equal to the work done on it by the electric field:

KE=qVKE = qV

Where:

  • V=6000extVV = 6000 ext{ V} is the potential difference.

Using kinetic energy: 12mv2=qV\frac{1}{2} mv^2 = qV

From which we derive: v=2qVmv = \sqrt{\frac{2qV}{m}}

Substituting in the values of charge (q=1.2imes1019extCq = 1.2 imes 10^{-19} ext{ C}) and mass (m=1.2imes1026extkgm = 1.2 imes 10^{-26} ext{ kg}) gives us the speed of the ion emerging through aperture P.

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 each ion to travel from aperture P to the detector can be related to their respective speeds, which are dependent on their masses:

  1. Ions with smaller masses will accelerate more and thus travel faster, leading to shorter time intervals to the detector.
  2. Conversely, heavier ions will have greater inertia, resulting in longer travel times.
  3. By measuring these time intervals and knowing the acceleration due to fields, the relative masses of the ions can be inferred based on the differences in their speeds.

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