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31% Sat 2:32 AM Q. Chrome File Edit View History Bookmarks People Window Help Rawan OC Chegg St Bbd Welcome G midag all G What is t 5. Plants 6. Plants MInbox (12 Fungi: Ne Fiverr r Google T Fiverr C fi https:// session masteringphysics.com 58831327 College Physics II PHYS245 Spring2016 Vemuru Signed in as rawan aljohani l Help l Clot h 20 HW Mass Spectrometer Resources 2 of 25 next previous Mass Spectrometer Part A J. J. Thomson is best known for his discoveries about the nature of After being accelerated to a speed o 1.98x105 m/s, the particle enters a uniform magnetic field of strength 0.800 T and travels in a circle of radius 30.0 cm (determined by cathode rays. His other important contribution was the invention, together observing where it hits the screen as shown in the figure). The results of this experiment allow one to find m/g with one of his students, of the mass spectrometer, a device that measures the ratio of mass m to (positive) charge q of an ion Find the ratio m/g for this particle The spectrometer consists of two regions as shown in the figure.(Figure Express your answer numerically in kilograms per coulomb. 1) In the first region an electric field accelerates the ion and in the second the ion follows a circular arc in a magnetic field. The radius of curvature of the arc can be measured and then the m/q ratio can be found m/g kg/C Submit Hint My Answers Give Up Revi incorrect; Try Again; 14 attempts remaining Figure 1 Continue Provide Feedback t Show All Pilobolus Data Spring....xlsx ecmr. docx electron charge.pdf ecmr (1) docx

Explanation / Answer

apply centripetal force = magnetic force

i.e mv^2/r = qvB,

so speed v = sqrt(2eV/m)

where m = mass o the charged particle

v = velocity, r = radius ,

q = charge = 1.6*10^-19 C

B = magnetic field

so now , so as r = mv/qB

so m/q = Br/v

m/q = (0.8 * 0.3)/(1.98 e 5)

m/q = 1.212*10^-6 kg /C

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