Mass Spectrometers
In 1913, J.J. Thomson at the Cavendish Laboratory, Cambridge, passed ionised neon through combined electric and magnetic fields and detected two distinct parabolas on a photographic plate — corresponding to masses 20 and 22 atomic mass units. This was the first detection of stable isotopes (Ne-20 and Ne-22), proving that atoms of the same element can have different masses. Thomson's parabola spectrograph is the direct ancestor of every modern mass spectrometer used in chemistry, biology, and forensic science.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A beam of positively charged ions enters a region of uniform magnetic field perpendicular to their velocity.
- Will the ions travel in a straight line, a circular path, or a parabolic path?
- Two ions have the same charge but different masses. Which one will curve more sharply — the lighter ion or the heavier ion?
- A mass spectrometer needs to separate ions by mass. Why is it important that all ions enter the magnetic field at the same speed?
Warm-up — in a mass spectrometer, the magnetic force on an ion acts as the ___ force that keeps it moving in a circular arc.
Know — Radius of Curvature
- A charged particle moving perpendicular to $B$ follows a circular path: $r = mv/(qB)$
- The magnetic force provides the centripetal force: $qvB = mv^2/r$
Understand — Mass Spectrometer Design
- A velocity selector uses crossed $E$ and $B$ fields to filter ions by speed: $v = E/B$
- 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 velocity selector works and why it is necessary
Core Content
The physics that makes mass spectrometry possible
Fire a stream of ions into a curved magnetic region and watch them emerge at different positions on a detector — heavy ions curving only slightly, light ions bending sharply. This is the mass spectrometer in action: the magnetic force on each moving ion acts as a centripetal force, curving the ion into a circular arc. The radius of that arc is directly proportional to the ion's mass — which is why ions of different masses land at different spots on the detector, allowing scientists to read off their masses from the geometry alone.
$r = \dfrac{mv}{qB}$
$r$ = radius (m) · $m$ = mass (kg) · $v$ = speed (m/s) · $q$ = charge (C) · $B$ = field strength (T)
Derivation: The magnetic force on a moving charge is $F = qvB$ (when $v \perp B$). This provides the centripetal force $F = mv^2/r$. Equating: $qvB = mv^2/r$. Solving for $r$: $r = mv/(qB)$.
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 $r$).
Figure 1 — Lighter ions curve more sharply than heavier ions at the same speed and charge
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 of curvature? Show your reasoning using $r = mv/(qB)$.
Derivation: $qvB = mv^2/r$ → $r = mv/(qB)$ (m). At equal charge and speed, lighter ions curve more sharply (smaller $r$); heavier ions curve less (larger $r$). $r \propto m/q$ — the ratio that mass spectrometers measure.
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:
Ensuring all ions enter the analyser at the same speed
We just saw that radius $r = mv/(qB)$ distinguishes masses — but only if all ions enter at the same speed. That raises a question: what device ensures every ion enters the analyser with exactly $v = E/B$? This card answers it → the velocity selector: crossed E and B fields that only pass ions where $qE = qvB$.
For a mass spectrometer to work, all ions must have the same speed when they enter the magnetic analyser. Otherwise, you couldn't tell whether a large radius is due to a heavy mass or a high speed.
A velocity selector solves this. 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 $F_B = qvB$
When these forces balance, the ion travels straight through:
$qE = qvB \quad \Rightarrow \quad v = \dfrac{E}{B}$
Only ions with speed $v = E/B$ pass through undeflected — mass and charge cancel out
Ions that are too slow are deflected by the electric field and hit the top plate. Ions that are too fast are deflected by the magnetic field and hit the bottom plate. Only ions with exactly $v = E/B$ emerge through the slit.
In velocity selector problems, the mass and charge of the ion cancel out — the selected speed $v = E/B$ is the same for all ions regardless of their mass or charge. This is why it works as a universal speed filter.
Velocity selector: $qE = qvB$ → $v = E/B$ (mass and charge cancel). Universal speed filter: any ion at $v = E/B$ passes undeflected; slower ions deflected by E-field; faster ions deflected by B-field.
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 $v = E/B$.
From ion source to detector — four stages
We just saw how the velocity selector provides all ions with the same speed. That raises a question: what is the full sequence from creating ions to measuring mass? This card answers it → ionise → accelerate → velocity-select → magnetically deflect; position on detector reveals $m/q$.
A complete mass spectrometer has four stages. Each stage performs a specific function to ensure that the final measurement depends only on mass-to-charge ratio.
- Ionisation: Atoms are ionised (electrons removed) so they can be accelerated and deflected by fields
- Acceleration: Ions are accelerated through a potential difference to give them kinetic energy
- Velocity selection: Crossed $E$ and $B$ fields filter ions to a single speed $v = E/B$
- Magnetic analysis: A uniform $B$-field bends ions into circular paths; radius reveals $m/q$
The detector records where each ion lands. Since $r = mv/(qB)$ and all ions have the same $v$ and $B$, the position depends only on $m/q$. This allows precise measurement of isotope masses.
In a mass spectrometer, singly ionised magnesium ions (charge $+e$) 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 isotope separation for Mg-26 ($m = 26$ u).
- Part (a) — speed. $v = E/B_1 = (5.0 \times 10^3)/(0.20) = 2.5 \times 10^4$ m/s
- Part (b) — Mg-24. $m = 24 \times 1.661 \times 10^{-27} = 3.99 \times 10^{-26}$ kg. $r_{24} = mv/(qB_2) = (3.99 \times 10^{-26})(2.5 \times 10^4)/[(1.602 \times 10^{-19})(0.40)] = 1.56 \times 10^{-2}$ m = 1.56 cm
- 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.
Mass spectrometer stages: ionise → accelerate → velocity-select ($v = E/B_1$) → analyse ($r = mv/qB_2$). Detector position reveals $m/q$. Separation after 180°: $\Delta x = 2\Delta r = 2(r_2-r_1)$. Worked result: $v = 2.5\times10^4$ m/s, $r_{24} = 1.56$ cm, separation Mg-24/26 = 0.26 cm.
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:
Use $r = mv/(qB)$ to explore how design choices affect isotope separation
- 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?
- Return to $m = 20$ u. Double the speed to $1.0 \times 10^5$ m/s. What happens to $r$? Explain why this shows the velocity selector is essential.
- 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.
Link mass spectrometry to earlier electromagnetism concepts
A student argues that a velocity selector is unnecessary because you could just accelerate all ions through the same potential difference, giving them all the same kinetic energy. Explain why this would not work for separating ions by mass. (Hint: consider how kinetic energy relates to speed and mass.)
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.
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): $m = 40 \times 1.661 \times 10^{-27} = 6.644 \times 10^{-26}$ kg. $r = mv/(qB) = (6.644 \times 10^{-26} \times 1.0 \times 10^5)/[(1.602 \times 10^{-19}) \times 0.80] = 6.644 \times 10^{-21}/(1.282 \times 10^{-19}) = 5.18 \times 10^{-2}$ m $\approx$ 5.2 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)
At the start you were asked about Thomson's 1913 parabola spectrograph at the Cavendish Laboratory: Ne-20 versus Ne-22, same charge and speed — which curves more sharply, and by what factor?
The answer: Ne-20 curves more sharply. $r = mv/(qB)$, so $r_{20}/r_{22} = 20/22 = 0.909$. Ne-20's radius is about 9% smaller — the two parabolas Thomson detected on his photographic plate were separated by this mass difference. This was the first experimental proof that isotopes of the same element exist with different masses.
The velocity selector ($v = E/B$) is independent of both mass and charge — without it, different ion speeds would produce overlapping arcs that make mass identification impossible.
Extend: A student claims the velocity selector could be replaced by simply accelerating all ions through the same potential difference. Using $KE = qV$ and $KE = \tfrac{1}{2}mv^2$, show algebraically why ions of different masses would still enter the analyser at different speeds, making mass separation ambiguous.
Mass spectrometry brings together multiple physics concepts:
- Magnetic force on moving charges ($F = qvB$) provides the centripetal force for circular motion
- Crossed electric and magnetic fields create a velocity selector: only $v = E/B$ passes through
- The radius of curvature $r = mv/(qB)$ 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 (not potential acceleration) is required to ensure speed uniformity