Question

Mass Spectrometry Part A: Match each statement with the term it most closely matches. A Hard...

Mass Spectrometry

Part A: Match each statement with the term it most closely matches.

A Hard ionization
B Electrospray Ionization (ESI)
C Molecular ion peak
D Ion-Trap
E Fast atom bombardment (FAB)
F Time of Flight (TOF)
G Base peak
H Quadrupole
I Soft ionization
J Matrix-Assisted Laser Desorption/Ionization (MALDI)
K Magnetic sector


1. ___ This ionization technique is used to analyze large, non-volatile molecules. The analyte is dissolved in a liquid matrix and placed on a target where it is impacted with an atom beam.
2. ___ Mass analyzer in which ions are separated by their velocity. Ions sequentially strike the detector in order of increasing m/z.
3. ___ This ionization technique is used to analyze large, non-volatile molecules. Ions are formed by placing the analyte in an excess of UV-absorbing molecules, followed by irradiation with a laser.
4. ___ This ionization technique is used to analyze large, non-volatile molecules. Ions are formed by injecting the sample solution across a high potential difference producing multiply protonated molecules whose charge tends to increase with the size of the molecule.
5. ___ General term used to describe ionization techniques that produces few ion fragments. Chemical ionization is an example of this.
6. ___ General term used to describe ionization techniques that produces many ion fragments. Electron ionization is an example of this.
7. ___ Analyzer that is based upon the deflection of ions in a magnetic field.

Part B:  A simple mass spectrometer may include an electron ionization (EI) source and magnetic sector mass analyzer. In this type of instrument, singly charged ions are produced and accelerated through the slit to the analyzer by applying high potentials to accelerator plates.

1. If an ion with mass 184 amu and charge z = 1 is accelerated by a potential of 6000 V, what is its kinetic energy (in J)?

Note: The following information may be useful for solving the two parts of this problem.

1 amu = 1.66 x 10-27 kg
Electronic charge, e = 1.602 x 10-19 C
1 J = 1 kg m2/s2
1 V = 1 J/C

2. What is the velocity of the ion?

0 0
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Answer #1

1- E

2- F

3- J

4- B

5- I

6- A

7- K

1 =

Kinetic energy = potential energy

lons accelerated by the electric field will acquire a kinetic energy equal to the potential energy of the electric field.

K. E = z * V

z=1e = 1.602*10^-19C

V= 6000 V = 6000 J/C

K.E = 1.602 *10^-19*6000 J

= 9.612*10^ - 16 J

2=

Velocity

K. E. = 1/2*mv^2

V = [(2*KE) / m] ^1/2

=[ (2*9.612*10^-16 kg m2/s2)/1.66 x 10-27 kg*184] ^1/2

=79333.92 m2/s2

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