Mass spectrometry has become an important measurement tool in clinical chemistry, microbiology, toxicology and in the pharmaceutical world. A mass spectrometer deflects ionized and accelerated molecular fragments using a magnetic field and sorts them according to their charge to mass ratio. For this exercise you will use the PASCO e/m apparatus like a mass spectrometer to determine the electron's charge to mass ratio, e/m, by measuring the radius of curvature of an electron's path in a uniform magnetic field of known strength.
|In the PASCO e/m apparatus a beam of electrons is
accelerated through a known potential difference, so the kinetic energy
and therefore the velocity of the electrons is known. A pair of
Helmholtz coils can produce a uniform magnetic field at right angles to
the electron beam. This magnetic field then deflects the electron beam
in a circular path. A unique feature of the e/m tube is that the socket
rotates, allowing the electron beam to be oriented at any angle (from
0-90 degrees) with respect to the magnetic field produced by the
Helmholtz coils. The vector nature of the magnetic force on moving
charged particles can therefore be explored. A small permanent magnet
can also be used to deflect the electron beam.
The Helmholtz Coils
The radius of the coils is equal to their separation. This geometry provides a highly uniform magnetic field near the center of the coils. The Helmholtz coils of the e/m apparatus have a radius and a separation of 15 cm. Each coil has 130 turns. The magnetic field B produced by the coils is proportional to the current I through the coils times 7.80*10-4 T/A. It is perpendicular to the plane of the coils.
B = (7.80*10-4 T/A) * I.
The control panel of the e/m apparatus is straightforward. All connections are labeled.
The hood can be placed over the top of the e/m apparatus so the experiment can be performed in a lighted room.
A mirrored scale is attached to the back of the rear Helmholtz coil. It is illuminated when the heater of the electron gun is powered. By lining the electron beam up with its image in the mirrored scale, the radius of the beam path can be measured without parallax error.
Experiment: Measuring e/m
Open a Microsoft Word document. Answer all the questions in blue font.
|Power supplies and meters are connected to the to the front panel of the e/m apparatus, as shown
in the figure below.|
|The power supplies are adjusted to the following levels:
|The current for the Helmholtz coils is adjusted with the current adjust knob. The ammeter displays the current in units of Ampere (A).|
|After the cathode has heated up, the electron
beam emerges from the electron gun. Its path is curved by the field from the
Helmholtz coils. The electron beam's path is parallel to the Helmholtz coils.
|Carefully read the current to the Helmholtz coils from the ammeter
(upper power supply) and the acceleration
voltage from the voltmeter (lower power supply) in the picture below.
Record the values in the table
|Carefully measure the radius of the circular path of the electron beam.
Look through the
tube at the electron beam. Measure the
radius of the path as you see it on both sides of the scale, then average the results.
(The markings on the scale are in units of cm.) Record your result in the table.
Paste your table into your Word document.|
The magnetic force Fm acting on a charged particle of charge q moving with velocity v in a magnetic field B is given by the equation Fm = qv × B.. Since the electron beam in this experiment is perpendicular to the magnetic field, we have the following equation relating the magnitudes Fm, q, v, and B.
Fm = qvB.
The electron is moving in a circular path of radius r, with the magnetic force being equal to the centripetal force mv2/r. We therefore have
qvB = mv2/r or q/m = v/Br.
We denote the magnitude of the charge q of the electron by e and therefore have
e/m = v/Br.
The electrons are accelerated through the accelerating potential V, gaining kinetic energy equal to their charge times the accelerating potential. Therefore eV =½mv2. The velocity of the electrons therefore is v = (2eV/m)1/2. Inserting this expression for v in the equation above we obtain
e/m = 2V/(Br)2.
Calculate e/m (in units of C/kg) from the measured values in your table.
Calculate e/m from the accepted values of the electron's charge and mass.
Do these two values agree? What is the percentage difference?
To earn extra credit add your name and e-mail address to your Word document.. Save your document (your name_exm7.docx), Go to Blackboard, Assignments, Extra Credit 7, and attach your document.