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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