AS Chemistry 9701: Time-of-Flight Mass Spectrometer Principles

In a time-of-flight (TOF) mass spectrometer, a sample is first ionised, then the resulting ions are accelerated through the same accelerating potential difference, allowed to drift along a flight tube of fixed length, and finally detected. The kinetic energy gained by an ion from this accelerating field depends on both the potential difference and the ion's own charge.

(a) Explain why the sample particles must be ionised before they can be accelerated by the electric field and detected. [2]

(b) An ion $\text{X}^+$ has mass number 40 and charge $1+$ . An ion $\text{Y}^{2+}$ has mass number 80 and charge $2+$ . Show that $\text{X}^+$ and $\text{Y}^{2+}$ have the same mass-to-charge ratio, and use this, together with the fact that $\text{Y}^{2+}$ gains twice the kinetic energy of $\text{X}^+$ from the accelerating field, to explain why the two ions take the same time to travel the length of the flight tube, even though a $\text{Y}^{2+}$ ion has twice the mass of an $\text{X}^+$ ion. [3]

(c) State and explain the effect on the time of flight of all the ions in the spectrometer if the accelerating voltage is increased. [1]

Part (a): Why ionisation is necessary

An electric field exerts a force only on a charged particle; a neutral atom experiences no net force from the field and would simply drift through the accelerating region unaffected, gaining no kinetic energy. Ionising the sample gives each particle a net electric charge, so the accelerating field can exert a force on it and speed it up. Ionisation also gives each particle the charge needed for the detector to register a signal (a small electric current) when the ion arrives.

Part (b): Mass-to-charge ratio and time of flight

The mass-to-charge ratio, $m/z$ , of each ion is:

$\left(\frac{m}{z}\right)_{\text{X}^+} = \frac{40}{1} = 40 \qquad \left(\frac{m}{z}\right)_{\text{Y}^{2+}} = \frac{80}{2} = 40$

So $\text{X}^+$ and $\text{Y}^{2+}$ have the same mass-to-charge ratio, $40$ .

Both ions are accelerated through the same potential difference, $V$ , but the kinetic energy an ion gains from this field depends on its own charge: $E_k = zeV$ , where $e$ is the charge on an electron and $z$ is the number of charges on the ion. Since $\text{Y}^{2+}$ carries twice the charge of $\text{X}^+$ , it gains twice the kinetic energy from the same accelerating field.

From $E_k = \frac{1}{2}mv^2 = zeV$ , the speed of an ion leaving the accelerating field is:

$v = \sqrt{\frac{2zeV}{m}} = \sqrt{\frac{2eV}{m/z}}$

Since $e$ and $V$ are the same for every ion, $v$ depends only on the ratio $m/z$ , not on $m$ and $z$ separately. Because $\text{X}^+$ and $\text{Y}^{2+}$ share the same $m/z = 40$ , they reach exactly the same speed leaving the accelerating field. $\text{Y}^{2+}$ has twice the mass of $\text{X}^+$ , but its double charge gives it double the kinetic energy, and these two factors exactly cancel in $v = \sqrt{2eV/(m/z)}$ , so the two ions reach the same speed and take the same time to cross a flight tube of fixed length.

Part (c): Effect of increasing the accelerating voltage

Increasing the accelerating voltage, $V$ , increases the kinetic energy, $E_k = zeV$ , gained by every ion, in proportion to its own charge $z$ . Since $v = \sqrt{2zeV/m}$ increases with $V$ for any ion, every ion travels faster through the flight tube, so the time of flight decreases for all ions (their relative arrival order, set by $m/z$ , is unchanged).

Final answers

(a) Only charged particles are accelerated by an electric field (and give a detectable current), so neutral atoms must first be ionised.
(b) Both ions have $m/z = \boxed{40}$ ; equal $m/z$ means equal speed once each ion’s own charge-proportional kinetic energy ( $E_k=zeV$ ) is accounted for, so $\text{X}^+$ and $\text{Y}^{2+}$ arrive together.
(c) Time of flight decreases for all ions, since every ion gains more kinetic energy (in proportion to its charge) and travels faster.