a short course of mass spectrometry

8/2/2019 A Short Course of Mass Spectrometry

http://slidepdf.com/reader/full/a-short-course-of-mass-spectrometry 1/34

1

All mass spectrometers consist of:

• A source of ions

• A mass analyser (or mass separator)

• A means of detecting ions

Ion sources

• Surface Ionisation

• Plasma Ionisation

• Electron Ionisation

• Secondary-Ion

• RIMS

Also called Thermal Ionisation



2

Ion sources




• Secondary-Ion

• RIMS

Ion sources




• Secondary-Ion

• RIMS



3

Ion sources




• Secondary-Ion

• RIMS

Also called Electron Impact Ionisation

Ion sources




• Secondary-Ion

• RIMS



4

Ion sources




• Secondary-Ion

• RIMSResonant (photo)ionisation

Other ion sources(not generally used for isotope ratio measurements)

• Glow Discharge

• Spark Source

• Electrospray

• Matrix-Assisted Laser Desorption Ionization (MALDI)

• Field Ionisation / Field Desorption

• Chemical Ionisation

• Fast Atom Bombardment (FAB) Ionisation

• …and no doubt many more



5

Mass Analysers

• Magnetic Sector

• Quadrupole

• Time-of-Flight (TOF)

• Fourier Transform

r = mv/qB

Mass Analysers

• Magnetic Sector

• Quadrupole





6

Mass Analysers

• Magnetic Sector

• Quadrupole



Mass Analysers

• Magnetic Sector

• Quadrupole





7

Mass Analysers

• Magnetic Sector

• Quadrupole



– Penning Trap

Mass Analysers

• Magnetic Sector

• Quadrupole


• Fourier Transform – Penning Trap

– Orbitrap



8

Detection

• Faraday Cup

• Secondary Electron

Multiplier

• Daly Detector

•

Gas Ionisation

Detection

• Faraday Cup


Multiplier

• Daly Detector • Gas Ionisation



9

Detection

• Faraday Cup


Multiplier

• Daly Detector

•

Gas Ionisation

Detection

• Faraday Cup


Multiplier

• Daly Detector • Gas Ionisation

– Energy and Z resolution



10

Finnigan Triton

• Thermal (surface)

ionisation

• Magnetic Sector, single

focussing

• Faraday, SEM, or

Channeltron detectors

Finnigan Neptune

• Inductively-coupled

plasma source

• Magnetic Sector, double

focussing• Faraday, SEM, or

Channeltron detectors



11

IMS 1270

• Secondary-Ion Source


focussing

• Faraday, or SEM

Shrimp RG



focussing

• Reverse Geometry• Faraday, or SEM



12

MegaSIMS – AMS (of sorts)



focussing

• Faraday, SEM



Charge and current

• Current: amp, A

• Charge: coulomb, C, equal to 1 amp for 1 second

•

Charge on the electron, e = −1.6× 10−19

C• Hence, number of electrons (or singly-charged ions) per second for current I is,

counts per second, cps = I/e

1

Electrostatic fields and potentials

• Charges give rise to the electrostatic field , E. For example, the field due to a point

positive charge has magnitude ∝ 1/r2.

• A charge, q, in an electrostatic field experiences a force, F, given by

F = qE

• Newton: work = force × distance traveled in the direction of the force.

• The work required to move a charge between two fixed points in an electrostatic field

is independent of the path taken . Such a field is called a conservative field.

• We call the work required to move unit charge from A and B the electrostatic potential

diff erence, ∆V , between points A and B.

• ∆V = W/q, where W is the work. With W in joules (J) and q in coulombs, V is in

volts.

• The electrostatic potential in a volume of space completely defines the electrostatic

field there. The direction of the field is perpendicular to lines of equipotential and

is exactly analogous to the line of steepest descent being perpendicular to contour

(equialtitude) lines.

2



. . . more electrostatic fields and potentials

• Note: the zero of potential is arbitrary. In practice it is often taken to be the potential

of the earth.

• A perfect electrical conductor is an equipotential. Why?

• A free charge will accelerate in the direction of the electrostatic field (F = ma).

• Regardless of the details of the (possibly complex) arrangement of fields, the kinetic

energy a charged particle gains traveling from A to B is q∆V . If the particle is initially

at rest we have

KE =1

2mv2 = q∆V

• The joule is an inconveniently large unit of energy to describe energies of individual ions

or electrons. Instead, we often use the electron-volt (eV) defined as the energy required

to move one electron through a potential diff erence of 1 volt. 1 eV ≈ 1.6×10−19 joules.

• The usual prefixes apply: keV, MeV . . .

• Tip: use mks units (metre, kilogram, second, [amp]) → energy in joules (J), charge in

coulombs (C), electrostatic potential in volts (V).

3

Example

1. A 6Li+ ion, initially at rest, is accelerated through a potential diff erence of 10 kV.

What is its final (a) kinetic energy in electron-volts (b) kinetic energy in joules (c) veloc-

ity. (Electronic charge = 1.6× 10−19 C, mass of proton and neutron = 1.7× 10−27 kg)

(a) 10 keV

(b) (1.6× 10−19 C)× (104 V) = 1.6× 10−15 joules

(c) Using KE = 1

2mv2

v2 = 2× 1.6× 10−15/(6× 1.7× 10−27) m2s−2

v = 5.6× 105 ms−1

2. What about the same problem applied to an electron? Electronic mass = 9 .1×10−31 kg.

(a) Same

(b) Same

(c) Using KE = 1

2mv2

v2 = 2× 1.6× 10−15/(9.1× 10−31) m2s−2

v = 5.9×

10

7

ms

−1

(Relativistic treatment gives v = 5.8×

10

7

ms

−1

).

4



Some applications of electrostatic fields in mass

spectrometry

• Acceleration of ions.

•

Beam steering. The field between (infinite) parallel plates is uniform and ⊥r

to plates:E ⊥ = ∆V /d.

• Electrostatic quadrupole lens (e.g. Neptune’s zoom quad) field is linear in x and y:

E x = ∆V x/a2, E y = −∆V y/a2, where 2a is the aperture of the lens

• Round and rectangular lenses.

• Cylindrical ESA: field is radial and ∝ 1/r,

E r =∆V

r ln(r2/r1)

• Spherical ESA: field is radial and ∝ 1/r2,

E r =∆V

r2(1/r1 − 1/r2)

5

E = (V2-V1) / {r ln(r2 /r1) } E = (V2-V1) / {r (1/r1 - 1/r2) } Ey = -2V0 y/ a

E

E

2r1

2r1

V2

V2

V1

V1

2r2

2r2

2 2

Ex = 2V0 x/ a2

E=V/d

+V/2

-V/2

d

6



Electrostatic analysers (ESAs)

• Consider a cylindrical or spherical ESA and an ion trajectory which passes exactly

mid-way between the plates. This is sometime called the central, median, or axial

trajectory.

• The electrostatic field E is constant along the ion’s path so the trajectory is circular.

Applying the formulae F = ma (Newton’s third law) and a = v2/r (acceleration due

to motion in a circle) we see that

qE = mv2/r

7

ESAs continued

• The radius of the trajectory, r, is proportional to mv2/q, the ion’s kinetic energy to

charge ratio.

• Electrostatic analysers are therefore energy analysers.

8



Momentum

• Momentum, p = mv.

• For ions, charge q, accelerated from rest though a potential diff erence V ,

12mv2 = qV

eliminating v gives, p2 = 2mqV

• Therefore, ions of diff erent mass but the same kinetic energy (i.e. the same accelerating

potential) have momentum ∝ √m.

• Why do we care about momentum?

9

Magnetic fields

• Motion of charged particles in magnetic fields: F = qv×B. The force acts in a plane

perpendicular to the plane defined by the directions of v and B. Since the force is

always perpendicular to the velocity, the field does no work on the charge, the kinetic

energy is unaff ected and the trajectory is helical if B is homogenous. In the special

case where v and B are at right angles, F = qvB and the trajectory is circular if B

is homogenous..

• Note: in standard SI units (mks system) B is in tesla: 1 T = 104 gauss. Earth’s field:

0.25–0.65 gauss.

• Radius, r, of circular trajectory follows directly from F = ma and F = qvB . For

circular motion, a = v2/r. Eliminating F and a gives,

mv2/r = qvB

r = p/qB

Hence the radius is proportional to the momentum to charge ratio. Magnetic fields

are momentum analysers.

10



Magnetic fields continued

• If ions are constrained (e.g. by slits) to follow a particular radius in order to be detected,

a given mass will be transmitted when

B =

p

qr

=

√2mqV

qr

=1

r

� 2mV

q

Remember: V is the potential diff erence used to accelerate the ion, m is the mass and

q is the charge. If V and r are fixed, B ∝ m/q.

11

Mass calibration

• We have seen that, for ions to follow a circular path of given radius, the mass vs field

relationship in a mass spectrometer is of the form B ∝ m/q. In an ideal system, this

would be the precise form of the mass calibration curve.

• In practice, the field is not perfectly homogeneous. For example, the field cannot

sharply drop to zero at the pole boundaries since this is not allowed by the governing

equations (Maxwell’s equations). Also, field control is imperfect since the field is onlymeasured in one small (∼ cm) region, not over the entire length of the magnet. Finally,

hysteresis ensures that the inhomogeneity will be dependent on the magnetic history

of the poles.

• Small and difficult to predict deviations from the ideal mass calibration curve are

therefore inevitable.

• The mass calibration must be determined experimentally with known masses (ions).

• Because of the hysteresis and temperature drift of the field sensor, the mass calibration

should be checked (i.e. peaks centred) frequently.

12



Field control

B

Electromagnet Coils

DAC18 bits

Digital to analogue converter

Vhall ∝ B

−

+

Vdac

Vout∝ (Vdac - Vhall)

Differential Amplifier Power Amplifier

B ∝Imag

Imag∝ (Vdac - Vhall)

Field (hall) probe

out

• Combining the relationships in the figure gives B ∝ V dac.

13



Geometric optics

• We only need to know the positions of the principal foci (F� and F��) and the focal

length, f.

F' F''F' F''

f

f f

p' p' p''f p''

Thin Lens Thick LensGoverning equation: p' p'' = f 2

z

x

y

• Image magnification is −f/p�.

• The term stigmatic is used to mean the simultaneous focusing in both the xz and yz

planes.

1

Focusing by magnetic sectors• The same thick-lens equations apply.

• The bend in the optic axis means that image and object spaces have separate coordi-

nate systems.

• f = r/ sinΦ. F� and F�� are a distance f cosΦ from the pole boundaries.

2



Focusing by magnetic sectors continued

• Mass dispersion: If the relative mass diff erence between two ions is ∆m/m (� 1)

and they are separated a distance d in the collector slit plane, the mass dispersion of

the spectrometer is md/∆m. The mass dispersion is constant, that is, independent of

mass.

3

Stigmatic Magnets• Note: in most modern instruments, pole boundaries are often not normal to the beam

path. Non-normal entry and exit strongly aff ects the focal properties in the radial plane

(as shown) but also in the axial plane. Axial focusing arises because of the combinded

eff ect of curvature of the field near the boundaries plus the non-normal indicence.

By this means, magnets can be stigmatically focusing. The diagram shows exactly

twice the image/object distances, relative to normal incidence, at equal entry/exit pole

angles of 12

sin−1(4/5) = 26.6◦.

4



Focusing by electrostatic fields

• Both spherical and cylindrical ESAs show similar focusing properties to magnetic

sectors.

• For a cylindrical sector, the formulae are: f = r/{√

2sin(√

2Φ)} and distance from

plate boundary to the principal foci given by f cos(√

2Φ).

• Spherical ESA’s are stigmatically focusing: f = r/ sinΦ with principal foci f cosΦ

from the plate boundary.

5

Double focusing instruments• The term double focusing is normally used to mean simultaneous spatial (or angular)

focusing and energy focusing. That is, in the focal plane (collector slit) ions of all

divergence angles and kinetic energies at the source slit are brought together. In

practice only a small range of angles and energies can be accurately focused.

Energy Dispersed Focus

Source

Collector

Spherical ESA

Magnetic Sector

Red and blue ions have the same mass but different energies.

Red have 10% more energy than blue

• Note that, as shown in the figure, the energies are only refocused in one plane; that of the collector slit. In this plane the instrument is said to be focusing achromatically (by

analogy with light optics). A fully achromatic instrument is possible (e.g. Cameca

ion microprobes) by the addition of a lens between ESA and magnet.

6



Peak shape and mass resolving power

Collector Slit

A B C

A

B

C

s

w

∆m

ws

s

5% full height

• More-or-less conventional definition of mass resolving power: m/∆m, where ∆m is

the full peak width at 5% of full peak height.

7

More mass resolving power• Single slit edge (Finnigan) definition of mass resolving power: m/∆m, where ∆m is

measured between 5% and 95% of peak height on a single edge. Definition OK if there

are only 2 masses to resolve.

∆m

5% full height

95% full height

8



Mass defect and nuclear binding energy

•

The diff

erence betweent the mass of the nucleus and the mass of the individual protonsand neutrons (×c2) is called the nuclear binding energy.

• All nuclei have diff erent binding energies and, hence, diff erent masses. This enables

isobars to be resolved by high mass resolving power.

• The mass defect of a nuclide or molecule is simply the diff erence between its mass in

unified atomic mass units (u) and the sum of all its protons and neutrons. Example:

Mass (u) Mass Defect (mu)1H 1.0078 +7.840

Ar 39.9624 −37.616O 15.9949 −5.140Ar16O 55.9573 −42.756Fe 55.9349 −65.1

Therefore, the 40Ar16O molecule is 22.4 mu heavier than 56Fe. The nominal mass

resolving power required to resolve these is ≈ 2500 (56/0.0224).

9

Binding Energy Curve

U235

U238

Fe56O16

C12

He4

Li6

Li7

He3

H3

H2

H1

Number of nucleons in nucleus

A v e r a g e b i n d i n g e n

e r g y p e r n u c l e o n ( M e V )

9

8

7

6

5

4

3

2

1

00 30 60 90 120 150 180 210 240 270

•56Fe has one of the most tightly bound nucleus. Fission of heavier species or fusion of

lighter species can release binding energy.

10



Abundance Sensitivity

1u

It

I0

• The abundance sensitivity (at a specified mass) is the ratio of the intensity of the tail

intensity 1 mass unit away to the peak intensity.

• The the tail intensity is a function of residual gas pressure in the vacuum. Ions

scattering off of gas molecules are deflected and may be detected in the spectrum far

from their correct mass.

• Ions lose energy if scattered therefore additional energy filtering before the detector

can discriminate against scattered ions. This is achieved by an additional ESA or, in

the Finnigan instruments, by a strong focusing lens, called the RPQ, which also actsas an energy filter.

• The RPQ’s transmission e ffi ciency for non-scattered ions is typically 70 − 80% but∼ 5% or less for scattered ions.

• The abundance sensitivity sets a limit on the maximum relative abundance two isotopes

(separated by 1u) at which the minor isotope can be detected.

11



07/01/2012

1

Zoom Optics

Zoom Optics



07/01/2012

2

Zoom Optics

Mass Focal Plane

• Mass focal plane not

perp. to axis

• Plane is curved, but

this is usually ignored

• Perfectly achromatic

only on axis

• Cups are angled to line

up with beam



07/01/2012

3

Faraday Cup Measurements• R is not precisely known. Need gain calibration using a very stable test current

generated electronically.

• Baseline ! 0, amp requires small input (bias) current. Doesnt matter if itis stable.

• Virtual amplifiers " reduces the gain calibration inaccuracy

• Cup factors. Very small effect.

• Thermal voltage (Johnson) noise across R (1 #) is Vn=(4kTR/t)1/2, t=measurementtime (s). Vn=41 µV s1/2 at T=300 K, R=1011$

• Long measurement: 30 min " Vn=1 µV.

• 1 ‰ (2 #) measurement would require 2 mV beam.

Ion counting: how its done

• Discriminator rejects small pulses and outputs

fixed-width (20 ns) logic-level pulses.



07/01/2012

4

Ion Counting: devil in the details• Not all output pulses are the same height. There is a pulse height

distribution.

• There are also noise pulses (possibly thermal electrons). These are

small and rejected by the discriminator. Some genuine, ion-induced

pulses are also rejected.

• The ratio of the number of ions in to the counted pulses out is the yield.

More details and devils

• The yield is measured frequently and can drift with time/age of the SEM.

• Darknoise (~baseline) can be extremely low: 1 count per minute.

• Near-simultaneous (<20 ns) ion arrivals cannot be distinguished. This is

the deadtime of the system.

• If f is the true count rate, ! the deadtime and f m the measured countrate :

f m" (1 - f m !) f

• Deadtime correction works well when f m ! << 1

• Non-linearity: a countrate-dependent effect (in addition to the effect of

deadtime) which must be empirically calibrated. Can be several permil

per decade. Possible causes: current drawn on SEM dynodes drops their

voltage, dynode heating,…?



07/01/2012

5

Faraday vs Ion Counting

Johnson Noise:

V=IR

Jn=(4RkT/t)1/2 (volts)

rel.Jn=(4kT/Rt)

1/2

/Irel.Jn∝(1/I) · (1/Rt)1/2

Shot Noise:

n=t · I/e

Sn=n1/2 (no. of ions)

rel.Sn=(1/n)

1/2

rel.Sn∝(1/I)1/2 · (1/t)1/2

Faraday vs Ion Counting

0.00001

0.0001

0.001

0.01

0.1

1

10

100

1000

0.0001 0.01 1 100 10000 R e l a t i v e N o

i s e ( 1!

‰ )

Beam Current (pA)

Johnson and Shot Noise

Measurement time 5 mins

1e11 ohms Jn

1e10 ohms Jn

1e12 ohms Jn

Shot Noise



07/01/2012

6

Publications

• H.E. Duckworth, R.C. Barber, and V.S. Venkatasubramanian. Mass

Spectrometry 2nd ed. Cambridge (1990).

• A. Septier. Focusing of Charged Particles. Vols I and II. Academic Press

(1967).

• M.E. Wieser and J.B. Schwieters. The development of multiple collector mass

spectrometry for isotope ratio measurements. Int J. Mass Spectrom. (2005)

242, 97-115

• J.M. Hayes and D.A. Schoeller. High precision pulse counting: liminationsand optimal conditions. Anal. Chem. (1977), 49(2), 306-311.

• A. Montaser. Inductively coupled plasma mass spectrometry. Wiley-VCH

(1998).



Chris Coath January 7, 2012 1

Please choose the best answer.

1. All mass spectrometers have

a) an accelerating potential diff erence of several megavolts

b) a magnetic field

c) a quadrupole lens

d) an ion detector

e) Faraday cups

2. Mass spectrometers are used in

a) organic chemistry

b) trace element measurementsc) isotope measurements

d) materials science

e) all of the above

3. An electrostatic field will always

a) accelerate ions in the direction of the field

b) accelerate ions in their direction of travel

c) accelerate ions perpendicular to their direction of travel

d) focus ions

e) both (a) and (d)

4. A magnetic field will always

a) accelerate ions in the direction of the field

b) accelerate ions in their direction of travel

c) accelerate ions perpendicular to their direction of travel

d) accelerate ions perpendicular to the direction of the field

e) both (c) and (d)

5. A 10 keV 40Ar2+ ion

a) travels at a speed in excess of 3×

108 m/s

b) travels at the same speed as a 10 keV 40Ar+ ion

c) travels faster than a 10 keV 40Ar+ ion by a factor of √

2

d) travels faster than a 10 keV 40Ar+ ion by a factor of 2

e) none of the above

6. A 10 kV potential diff erence accelerates 238U+ ions from rest. Their finalspeed is

a) 89 km/s

b) 9.2 m/s




c) 63 km/s

d) 2.8 km/se) 84 m/s

7. A beam of ions are detected using a Faraday cup. The Faraday cupamplifier has a 1012 ohm feedback resistor. If the output of the amplifieris 6 volts, the ion beam current is

a) 6 pA

b) 60 pA

c) 600 pA

d) 6 nA

e) 60 nA8. The beam in question (7) is identified as 238U+ ions. A simultaneous mea-

surement of 234U+ is made on the secondary-electron multiplier. Ignoringany mass-bias and given an SEM yield of 98%, a transmission efficiencythrough the RPQ of 80%, and 234U/238U= 5.29× 10−5 the count rate, incounts per second (cps), on the SEM is

a) 2530 cps

b) 159 cps

c) 1980 cps

d) 1560 cps

e) 1590 cps

9. What is the greatest relative precision (1σ) which could be expected of the 234U/238U ratio measured in question (8) for a measurement taking10 minutes?

a) 64 ppm

b) 0.1 ppm

c) 25 �

d) 8 �

e) 1 �

10. The same U isotope beams in question (8) are nowboth

measured byFaraday cup, the minor isotope beam being amplified using a 1011 ohmfeedback resistor. Assuming an ambient temperature of 300 K, what rel-ative precision (1σ) could be achieved in a 10 minute measurement?

a) 100 �

b) 50 �

c) 20 �

d) 200 �

e) 1 �




11. What (nominal) mass resolving power is required to resolve 40Ar4+ (mass

defect -37.6 mu) from10

B+

(mass defect +12.9 mu)?a) 0.45

b) 450

c) 1800

d) 2900

e) 2.2

12. What (nominal) mass resolving power is required to resolve 40Ar12C+

from 52Cr+ (mass defect -59.5 mu)?

a) ∞

b) 540c) 2400

d) 1800

e) 420

a short course of mass spectrometry

Documents