Question

28: A limitation on how many spectra per second can be recorded by a time-of-flight mass spectrometer is the time it takes the slowest ions to go from the source to the detector. Suppose we want to scan up to m/z 500. Calculate the speed of this heaviest ion if it is accelerated through 5.00 kV in the source. How long would it take to drift 2.00 m through a spectrometer? At what frequency could you record spectra if a new extraction cycle were begun each time this heaviest ion reached the detector? What would be the frequency if you wanted to scan up to m/z 1000?

Short answer

The speed of an ion$v=4.3920\times {10}^4\frac{m}{s}$

Time of the heaviest particle T= $4.5558\times {10}^{-5}\mathrm{seconds}$

Frequency of the heaviest particle f= $=2.1950\times {10}^4{s}^{-1}$

Velocity of the 1000 Da ion v= $3.1056\times {10}^4\frac{m}{s}$

The time of the Da ion t= $6.439979392\times {10}^{-5}s$

Frequency of the 1000 Da ion f=$15528s^{-1}\approx 1.6\times {10}^4{s}^{-1}$

Step-by-step solution

Mass Spectrometer

Mass spectrometry is an analytical technique that is used to measure the mass-to-charge ratio of ions. The results are presented as a mass spectrum, a plot of intensity as a function of the mass-to-charge ratio.

Speed of the heaviest ion

To compute for the speed of an ion, equation 1 is used:

$v=\sqrt{\frac{2zeV}{m}}$

The equation should be first converted into kg,

$500Da\times \frac{1.661\times {10}^{-27}kg}{1Da}=8.305\times {10}^{-25}kg$

Then,

$$\begin{gathered} v=\sqrt{\frac{\left(2\right)\left(1\right)(1.602\times {10}^{-19}C)(5.00\times {10}^{3}V)}{(8.305\times {10}^{-25}kg)}} \\ =4.391990959\times {10}^4\sqrt{\frac{CV}{kg}}=4.3920\times {10}^4\frac{m}{s} \end{gathered}$$

Determining the time and frequency:

To determine the time needed for the heaviest particle to travel 2.00 m

$\frac{2.00m}{4.3920\times {10}^4{\displaystyle \frac{m}{s}}}=4.5558\times {10}^{-5}seconds$

Determining the frequency;

$$\begin{gathered} f=\frac{1}{4.5558\times {10}^{-5}seconds} \\ =2.1950\times {10}^4{s}^{-1} \end{gathered}$$

The difference between the velocities of the ions with 500 Da and 1000 Da is a factor of role="math" localid="1663659107041" $\frac{1}{\sqrt{2}}$.

Velocity of the 1000 Da ion,

$4.3920\times {10}^4\times \frac{1}{\sqrt{2}}=3.1056\times {10}^4\frac{m}{s}$

Using the speed of the Da ion, time needed for one cycle:

$t=\frac{2.00m}{3.1056\times {10}^4{\displaystyle \frac{m}{s}}}=6.439979392\times {10}^{-5}s$

Frequency of the 1000 Da ion:

$f=\frac{1}{6.439979392\times {10}^{-5}s}=15528s^{-1}\approx 1.6\times {10}^4{s}^{-1}$

The final answer is:

The speed of an ion $v=4.3920\times {10}^4\frac{m}{s}$

Time of the heaviest particle T= $4.5558\times {10}^{-5}seconds$

Frequency of the heaviest particle f= $=2.1950\times {10}^4{s}^{-1}$

Velocity of the 1000 Da ion v= $3.1056\times {10}^4\frac{m}{s}$

The time of the Da ion t=$6.439979392\times {10}^{-5}s$

Frequency of the 1000 Da ion f=$15528s^{-1}\approx 1.6\times {10}^4{s}^{-1}$