Year 12 Physics Module 6 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 7 of 21 IQ1: Charged Particles in Fields

Mass Spectrometers

Build the complete instrument from ion source to detector, compare two valid ways to prepare the beam, and use magnetic curvature to measure mass-to-charge ratio.

Today's hook: Two singly charged neon isotopes enter the same uniform magnetic analyser at the same perpendicular speed. Which curves more sharply, and by what factor? Use $r=mv/(|q|B)$.
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1
Retrieve the field physics and map what a mass spectrometer actually measures.
Beyond the syllabus. This whole lesson is an extension application: mass spectrometer design, isotope separation and instrument calculations are not named Module 6 requirements. The magnetic-radius physics it reuses is core — practise $r = \frac{mv}{qB}$ in its core setting, and read this lesson as enrichment that will not affect your exam readiness.
Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Use the available foundations worksheet for guided practice on this lesson.

Build Foundations & guided practice
Before you read, predict

A beam of positively charged ions enters a region of uniform magnetic field perpendicular to their velocity.

  1. Will the ions travel in a straight line, a circular path, or a parabolic path?
  2. Two ions have the same charge but different masses. Which one will curve more sharply, the lighter ion or the heavier ion?
  3. In a velocity-selected design, why must admitted ions enter the analyser at the same known speed? Can you think of another measurable way to prepare the beam?

Warm-up, in a mass spectrometer, the magnetic force on an ion acts as the ___ force that keeps it moving in a circular arc.

Learning Intentions
goals

Know, Radius of Curvature

  • A charged particle moving perpendicular to $B$ follows a circular path of radius $r=mv/(|q|B)$
  • For perpendicular entry, $|q|vB=mv^2/r$ in magnitude

Understand, Mass Spectrometer Design

  • One design uses crossed $E$ and $B$ fields to filter ions by speed: $v = E/B$
  • Another design accelerates ions from rest through a known potential difference
  • The magnetic analyser separates ions by mass-to-charge ratio
  • Lighter ions curve more sharply; heavier ions curve less

Can Do, Analyse and Calculate

  • Calculate the radius of curvature for an ion given $m$, $v$, $q$, and $B$
  • Predict the separation between two isotopes in a spectrometer
  • Explain how a crossed-field velocity selector works and when that design is useful
Scan these before reading
vocab
Mass spectrometerAn instrument that separates ions by mass-to-charge ratio using electric and magnetic fields.
Velocity selectorRegion of crossed $E$ and $B$ fields where only ions with $v = E/B$ pass through undeflected.
Mass-to-charge ratio ($m/|q|$)The magnitude ratio inferred from analyser curvature. Charge sign determines the side of deflection.
IsotopesAtoms of the same element with different numbers of neutrons, different masses, same charge.
Centripetal forceThe inward force needed to keep an object moving in a circle: $F = mv^2/r$.
Cross-lesson links: L04 established magnetic force, L05 derived circular motion, and L06 derived crossed-field selection. L18 now owns the complete instrument, detector geometry and comparison with known-potential acceleration.

Core Content

2
Derive analyser radius and separate magnitude from curvature direction.
1
Charged Particles in Magnetic Fields
+5 XP

The physics that makes mass spectrometry possible

Fire a stream of ions into a uniform magnetic region and they emerge at different detector positions. For perpendicular entry the magnetic force provides centripetal acceleration, with radius $r=mv/(|q|B)$. Radius therefore reveals mass-to-charge ratio only when the entry-speed preparation and charge state are known; geometry alone does not identify mass independently of $v$ and $|q|$.

Radius of curvature

$r = \dfrac{mv}{|q|B}$

$r$ = radius (m) · $m$ = mass (kg) · $v$ = speed (m/s) · $q$ = charge (C) · $B$ = field strength (T)

Derivation: For $v\perp B$, the magnetic-force magnitude is $F_B=|q|vB$. Equating this with $mv^2/r$ gives $|q|vB=mv^2/r$, so $r=mv/(|q|B)$. The sign of $q$ determines which way the path curves.

This equation reveals the core principle of mass spectrometry: for ions with the same charge and speed, the radius depends only on mass. Lighter ions curve more sharply (smaller $r$); heavier ions curve less (larger $r$).

Positive ions following circular arcs in a magnetic field Two positive ions enter to the right at the same speed in a uniform magnetic field directed into the page. Their paths curve upward. The lighter ion follows the smaller-radius circular arc and the heavier ion follows the larger-radius circular arc. Heavy ion, larger r Light ion, smaller r + ions, v B into page

Figure 1, positive ions at equal speed and charge magnitude follow circular arcs; the lighter ion curves more sharply.

Stop & check

An ion with mass $2m$ and charge $+q$ enters a magnetic field at speed $v$. Another ion with mass $m$ and charge $+2q$ enters at the same speed. Which ion has the larger radius? Show your reasoning using $r=mv/(|q|B)$.

Magnitude derivation: $|q|vB=mv^2/r$ → $r=mv/(|q|B)$. At equal speed and charge magnitude, lighter ions curve more sharply. The analyser measures $m/|q|$ when speed preparation is known; charge sign selects the side of deflection.

Pause, copy the highlighted radius formula and separation principle into your book before moving on.

An ion with mass $m$ and charge $q$ enters a magnetic field $B$ at speed $v$ perpendicular to the field. Its radius of curvature is $r$. If the field strength is doubled (and all other quantities stay the same), the new radius is:

3
Use vector force cancellation to explain an optional velocity selector.
2
The Velocity Selector
+5 XP

Preparing a common known speed in one instrument design

We just saw that $r=mv/(|q|B)$ depends on speed as well as $m/|q|$. That raises a question: how can one design control the entry speed? This card answers it → crossed E and B fields can pass ions with $v=E/B$. This is useful but optional; known-potential acceleration is another valid preparation.

A magnetic analyser needs the entry-speed preparation to be known. One option is to admit only ions with the same speed, so a radius difference cannot be confused with a speed difference.

A velocity selector provides this common-speed beam. It consists of crossed electric and magnetic fields:

  • The electric field pushes positive ions upward with force $F_E = qE$
  • The magnetic field pushes positive ions downward with force $F_B = qvB$

For positive ions moving to the right, choose $\vec E$ upward and $\vec B$ out of the page. Then $q\vec E$ is upward while $q(\vec v\times\vec B)$ is downward. When these vectors cancel, the ion travels straight through:

Velocity selector condition

$|q|E = |q|vB \quad \Rightarrow \quad v = \dfrac{E}{B}$

Only ions with speed $v = E/B$ pass through undeflected, mass and charge cancel out

In this stated orientation, ions that are too slow deflect upward because the electric force dominates; ions that are too fast deflect downward because the magnetic force dominates. Only ions with exactly $v = E/B$ emerge through the slit. Reversing the charge reverses both force vectors, so the same speed condition remains.

Crossed-field velocity selector for positive ions Positive ions move right between horizontal plates. The electric field and electric force point up. The magnetic field points out of the page, making the magnetic force point down. Equal and opposite forces select speed E divided by B. negative plate positive plate E, FE + ion, v FB B out of page (dots)

Figure 2, a stated vector orientation matters. The selector condition follows only when the electric and magnetic forces oppose.

Key insight

In velocity-selector problems, the charge magnitude cancels, so the selected speed $v=E/B$ is the same for all admitted ions regardless of mass or charge magnitude.

Velocity selector: opposing force magnitudes satisfy $|q|E=|q|vB$, so $v=E/B$. This speed condition is independent of mass and charge magnitude, but the force directions must be checked from $\vec E$, $\vec v$, $\vec B$ and the charge sign.

Add the highlighted velocity selector equation and filtering principle to your notes before the check below.

In a velocity selector, the electric and magnetic forces on the selected ion are equal in magnitude and opposite in direction.

The speed selected by a velocity selector depends on the mass of the ion passing through.

A heavier ion and a lighter ion (same charge) both pass through the velocity selector undeflected if they have the same speed $v = E/B$.

4
Derive a second valid design using acceleration through a known potential difference.
3
Known-Potential Acceleration
+5 XP

Different speeds can still produce a measurable mass-to-charge separation

We just saw one way to prepare a common speed. That raises a question: is a selector necessary? This card answers it → no. If ions start from rest and pass through a known accelerating potential, their speeds differ with mass, but the analyser radius still depends predictably on $m/|q|$.

Energy preparation

Let $\Delta V_{\mathrm{acc}}$ denote the positive magnitude of the accelerating potential change. For an ion starting from rest, the field gives it kinetic energy

$\Delta K=|q|\Delta V_{\mathrm{acc}}=\tfrac12mv^2$.

This magnitude form avoids confusing the positive energy gain with the signed relation $\Delta K=-q(V_f-V_i)$.

Magnetic analysis

Substitute $v=\sqrt{2|q|\Delta V_{\mathrm{acc}}/m}$ into $r=mv/(|q|B)$:

$r=\dfrac{1}{B}\sqrt{\dfrac{2m\Delta V_{\mathrm{acc}}}{|q|}}$.

Therefore $r^2=2m\Delta V_{\mathrm{acc}}/(|q|B^2)$, so detector position still separates $m/|q|$.

Do not overclaim

Equal accelerating potential does not give different masses the same speed. That does not make the design fail. It gives a different calibrated radius law. In either design, charge state must be known before $m/|q|$ can be converted to isotope mass.

Two valid preparations: a crossed-field selector gives $v=E/B$ and hence $r\propto m/|q|$ at common speed; known-potential acceleration gives $r=(1/B)\sqrt{2m\Delta V_{\mathrm{acc}}/|q|}$ and hence $r^2\propto m/|q|$.

Singly charged ions start from rest and cross the same known accelerating potential before entering the same analyser field. If one ion has four times the mass of another, its analyser radius is:

5
Trace the source-to-detector sequence and connect radius to detector separation.
4
The Complete Mass Spectrometer
+5 XP

From ion source to detector, with an optional selector branch

We just compared two beam-preparation designs. That raises a question: what is the full sequence from creating ions to measuring mass? This card answers it → ionise → prepare a known beam → magnetically analyse → detect; position reveals $m/|q|$ under the chosen calibration.

A complete mass spectrometer needs four functions. The hardware can implement the preparation stage in different ways, so a crossed-field selector is not a universal requirement.

  1. Ionisation: Atoms are ionised (electrons removed) so they can be accelerated and deflected by fields
  2. Beam preparation: Slits make a narrow beam; the design either accelerates then selects $v=E/B$, or accelerates from rest through a known positive magnitude $\Delta V_{\mathrm{acc}}$
  3. Magnetic analysis: A uniform field perpendicular to entry bends ions into circular paths
  4. Detection: Detector position records the diameter $2r$ in a 180° analyser and therefore the calibrated $m/|q|$

For two landing points on the same side of the entrance slit, the separation magnitude is $s=|2r_2-2r_1|=2|r_2-r_1|$. This geometry result assumes both ions complete semicircular paths in the same analyser field.

Complete mass spectrometer with two beam-preparation options A positive ion source feeds a narrow beam. One branch accelerates the ions then uses crossed electric and magnetic fields to select a common speed. A second branch accelerates ions from rest through a known positive potential-difference magnitude. Either branch enters to the right into a uniform magnetic analyser directed into the page, where positive ions follow right-side semicircles and strike different positions on the detector line. positive-ion source narrow beam and slit accelerate + selector v = E/B known potential ΔK = |q|ΔVacc detector peak 2, diameter 2r₂ detector peak 1, diameter 2r₁ uniform magnetic analyser B into page, positive ions

Figure 3, two valid preparation branches feed a 180° analyser. For the stated positive-ion and field directions, the right-side semicircles return to the detector line.

Explore magnetic force and analyser curvature Open fullscreen ↗

Use the magnetic-force mode to test how $m$, $|q|$, $v$ and $B$ affect curvature. Treat the selector and detector drawing above as the canonical complete-instrument diagram.

Worked example, isotope separation

In a mass spectrometer, singly ionised magnesium ions (charge magnitude $|q|=e$) pass through a velocity selector with $E = 5.0 \times 10^3$ V/m and $B_1 = 0.20$ T. They then enter a uniform magnetic field of $B_2 = 0.40$ T perpendicular to their velocity. Calculate (a) the ion speed, (b) the radius for Mg-24 ($m = 24$ u), and (c) the detector separation for Mg-26 ($m = 26$ u) after 180° deflection.

  1. Part (a), speed. $v = E/B_1 = (5.0 \times 10^3)/(0.20) = 2.5 \times 10^4$ m/s
  2. Part (b), Mg-24. $m = 24 \times 1.661 \times 10^{-27} = 3.99 \times 10^{-26}$ kg. $r_{24} = mv/(|q|B_2) = (3.99 \times 10^{-26})(2.5 \times 10^4)/[(1.602 \times 10^{-19})(0.40)] = 1.56 \times 10^{-2}$ m = 1.56 cm
  3. Part (c), Mg-26. $m = 26 \times 1.661 \times 10^{-27} = 4.32 \times 10^{-26}$ kg. $r_{26} = (4.32 \times 10^{-26})(2.5 \times 10^4)/[(1.602 \times 10^{-19})(0.40)] = 1.69$ cm. Separation $= 2(r_{26} - r_{24}) = 2(1.69 - 1.56) = 0.26$ cm.

One design: ionise → accelerate → velocity-select ($v=E/B_1$) → analyse with $r=mv/(|q|B_2)$. Detector position reveals $m/|q|$. For 180° geometry, separation magnitude is $2|r_2-r_1|$. A selector is optional; known-potential acceleration is another valid design.

Pause, write the highlighted four stages and separation formula into your book before moving on.

A velocity selector has $E = 2.0 \times 10^4$ V/m and $B = 0.50$ T. The speed of ions that pass through undeflected is:

Resolution and uncertainty

For a 180° analyser, detector separation is $2|r_2-r_1|$. Peaks are distinguishable only when this separation exceeds broadening from slit width, detector resolution, field non-uniformity, spread in entry speed and uncertainty in $B$, $E$ or $\Delta V_{\mathrm{acc}}$. The model assumes perpendicular entry, a uniform analyser field, non-relativistic ions, negligible collisions in a sufficiently good vacuum and negligible interactions within the dilute beam. Report the measured quantity as $m/|q|$ unless charge state is independently known.

6
Calculate isotope separation, compare designs and evaluate resolution limits.
Activity 1, Radius and Separation Calculations
ApplyBand 4

Use $r=mv/(|q|B)$ to explore how design choices affect isotope separation

  1. An ion with mass $20$ u and charge $+e$ moves at $v = 5.0 \times 10^4$ m/s in $B = 0.50$ T. Calculate $r$. Now double the mass to $40$ u. What happens to $r$? By what factor does it change?
  2. Return to $m = 20$ u. Double the speed to $1.0 \times 10^5$ m/s. What happens to $r$? Explain why the analyser needs a calibrated beam preparation, while noting that a crossed-field selector is only one way to provide it.
  3. Imagine two isotopes: mass $20$ u and $22$ u, both with charge $+e$ and speed $5.0 \times 10^4$ m/s. Calculate their radii in $B = 0.50$ T, and determine how far apart they land on the detector after 180° deflection.
Activity 2, Concept Connections
UnderstandBand 4

Link mass spectrometry to earlier electromagnetism concepts

Compare two valid analyser designs: ions selected to a common speed by crossed fields, and ions accelerated from rest through a known positive potential-difference magnitude $\Delta V_{\mathrm{acc}}$. Derive how the magnetic-analysis radius depends on $m/|q|$ in each design and state the non-relativistic, uniform-field assumptions.

7
Retrieve the shared bank, then calculate radius and detector separation.
Quick recall, mass spectrometers
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 7 marks
+5 XP

ApplyBand 4(2 marks) 1. An ion with mass $40$ u and charge $+e$ moves at $1.0 \times 10^5$ m/s perpendicular to a magnetic field of $0.80$ T. Calculate the radius of curvature of the ion's path. (Give your answer in centimetres.)

1 mark: correct substitution with $m$ converted to kg · 1 mark: correct answer ~5.2 cm

AnalyseBand 5(5 marks) 2. In a mass spectrometer, singly ionised neon ions pass through a velocity selector with $E = 8.0 \times 10^3$ V/m and $B_1 = 0.40$ T. They then enter a magnetic field $B_2 = 0.60$ T perpendicular to their velocity.

  • Calculate the speed of ions emerging from the velocity selector. (1 mark)
  • Calculate the radius of curvature for Ne-20 ions (mass 20 u). (2 marks)
  • Ne-22 ions (mass 22 u) enter with the same speed. Calculate their radius and determine the linear separation between Ne-20 and Ne-22 on the detector after 180° deflection. (2 marks)

1 mark: correct $v = E/B_1$ · 1 mark each: correct $r_{20}$ and $r_{22}$ with working · 1 mark: correct separation = 2Δr

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

SAQ 1 (2 marks): $m=40(1.661\times10^{-27})=6.644\times10^{-26}$ kg. $r=mv/(|q|B)=(6.644\times10^{-26}\times1.0\times10^5)/[(1.602\times10^{-19})(0.80)]=5.18\times10^{-2}$ m $\approx5.2$ cm. (2 marks)

SAQ 2 (5 marks): (a) $v = E/B_1 = 8.0 \times 10^3/0.40 = 2.0 \times 10^4$ m/s. (1 mark) (b) $m_{20} = 20 \times 1.661 \times 10^{-27} = 3.322 \times 10^{-26}$ kg. $r_{20} = (3.322 \times 10^{-26} \times 2.0 \times 10^4)/[(1.602 \times 10^{-19}) \times 0.60] = 6.91 \times 10^{-3}$ m $= 0.691$ cm. (2 marks) (c) $m_{22} = 3.654 \times 10^{-26}$ kg. $r_{22} = (3.654 \times 10^{-26} \times 2.0 \times 10^4)/(9.612 \times 10^{-20}) = 7.60 \times 10^{-3}$ m $= 0.760$ cm. Separation $= 2(r_{22} - r_{20}) = 2(0.760 - 0.691) = 0.138$ cm $\approx 0.14$ cm. (2 marks)

8
Use the repaired review pool, revisit the isotope hook and state model limits.
Check what actually stuck
How did your thinking change?

At the start you were asked about Ne-20 versus Ne-22 at the same charge magnitude and speed: which curves more sharply, and by what factor?

The answer: Ne-20 curves more sharply. For equal $v$, $B$ and charge magnitude, $r=mv/(|q|B)$, so $r_{20}/r_{22}=20/22=0.909$. Ne-20's circular radius is about 9% smaller.

A crossed-field velocity selector fixes $v=E/B$, independent of ion mass and charge, so the analyser radius directly measures $m/|q|$. It is useful, but it is not required for every mass spectrometer design.

Extend: Suppose the selector is replaced by acceleration from rest through a known positive magnitude $\Delta V_{\mathrm{acc}}$. Combine $|q|\Delta V_{\mathrm{acc}} = \tfrac{1}{2}mv^2$ with $r=mv/(|q|B)$ to show how the analyser radius still separates ions by mass-to-charge ratio. State the uniform-field and non-relativistic conditions.

Synthesis, Connect the Ideas

Mass spectrometry brings together multiple physics concepts:

  • Magnetic-force magnitude $F_B=|q|vB$ provides centripetal force for perpendicular entry
  • Crossed electric and magnetic fields create a velocity selector: only $v = E/B$ passes through
  • The radius $r=mv/(|q|B)$ reveals mass-to-charge ratio when $v$ and $B$ are known
  • Applications include isotope dating, forensic analysis, drug testing, and protein identification
  • A velocity selector is optional: crossed fields give $v=E/B$, while known-potential acceleration gives $r=(1/B)\sqrt{2m\Delta V_{\mathrm{acc}}/|q|}$, where $\Delta V_{\mathrm{acc}}$ is a positive magnitude
← Previous: Electric vs Magnetic Fields Next: The Motor Effect →

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