Transcript Document

Mass spectrometer
Problem review
• An electron moving perpendicular to a
magnetic field of 4.60 x 10-3 T follows a
circular path of radius 2.80 mm.
• What is the electron’s speed?
• e = 1.6 x 10-19 c
• mass of e = 9.11 x 10-31 kg
An electron moving perpendicular to a magnetic field of 4.6x10-3
T follows a circular path of radius 2.80 mm.
What
is
the
electron’s
speed?
• Solving for v
• r = (m v ) /(q B)
• v = ( r q B) / m
• = (2.8x10-3 m)( 1.6x10-19 c)(4.6x10-3 T) /
(9.11x10-3kg)
v =(2.26x106 m/s)
Summary review
• Magnetic force can supply centripetal force
and cause a charged particle to move in a
circular path of radius
• r = (m v ) /(q B)
• where v is the component of the velocity
perpendicular to B for a charged particle with
mass m and charge q
REVIEW
Review
Magnetic fields can be used to
separate ISOTOPES
• mass spectrometer • Determine which
instrument which can
chemical elements go
measure the masses
into a sample you’re
and relative
analyzing
concentrations of atoms
and molecules.
• rely on orbit in
magnetic filed
• Makes use of circular
motion in a magnetic
filed to separate
isotopes.
Mass Spectrometer
Ratio of q/m
• an ion's path curves depends on two factors:
the mass of the ion and its charge:
• q/m= 2V / B2r2
•
Importance - two particles with the q/m ratio move in the same
path in a vacuum when subjected to the same electric and magnetic
fields.
• Its SI units are kg/C
Mass spectroscope finding the mass
• Given r , q , B , solve for m
( mass)
• The accelerating electric
potential V gives each
ionized atom a kinetic
energy :
• qV = ½ mv2
• Solve for velocity v :
• v = ( 2qV / m)1/2
• The radius of curvature:
• r = (mv)/(qB)
• Solve for m
• m= (rqB) / v
• Found v to be v = ( 2qV /
m)1/2 sub for v:
• m = rqB / ( 2qV /m)1/2
• Square both sides
• m2= r2 q2 B2 / (eqV/m)
• m = (qr2/2V) B2
Applications of Mass Spectrometers
• Carbon dating and
radioactive dating
process
• Detection of trace
quantitates of
contaminants or toxins
• Analyze the solar wind
• Two isotopes of uranium; U-235 and U-238
are sent into a mass spectrometer with a
speed of 1.05 x 105 m/s. Given the mass of
each isotope ( mu-235 = 3.90x10-25 kg and mu-25 kg) the strength of the
=
3.95x10
238
magnetic filed ( B = 0.750 T) and the charge of
each isotope ( q = 1.6x10-19 C ) find the
distance, d, between the tow isotopes after
they complete half of a circular path.
Known
• mu-235 = 3.90x10-25 kg mu-25 kg
=
3.95x10
238
• v = 1.05 x 105 m/s
• q = 1.6 x 10-19 C
• B = .750 T
• d=?
• 1- Draw a picture
• 2- determine the radius of the
circular path of the mu-235
• 3- determine the radius of the
circular path of the mu-235
• 4-calculate the separation
between the isotopes.
• R-mu-235 =( (3.90x10-25 kg)
( 1.05x105m/s)) / ((
1.6x10-19)(.750 T))=.341 m
• Rmu-238= ((3.95x10-25 kg)(
1.05x105m/s)) / (( 1.6x1019)(.750 T))= .346
• 2(.346 m -.341 m ) = .01m