A neon atom of mass 3.82×10-26 kg is triplyionized. That is, three electrons are

Question

A neon atom of mass 3.82×10-26 kg is triplyionized. That is, three electrons are stripped from the atom sothat its charge is +3e, or 4.80×10-19 C.The atom is accelerated across a potential difference of magnitude212 V, and enters a mass spectrometer whose magnetic field hasa strength of 1.25 T.

(a) What is the radius of the atom's trajectory in themagnetic field?
(b) What is the radius of the trajectory of a sulfur ion, alsotriply ionized, but having mass5.64×10-26 kg?
(c) What is the distance between the points on the detectorplate struck by these two atoms?

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Explanation / Answer

a) Charge of the triply ionized atom is q=+3e=4.80*10-19 C     Now the atom is accelerated through the potential difference V=212Volts     Then work done to accelerate the charge is equal to the change in the kinetic energy of the ion Then q(V)=(1/2)mv2 Here the mass of the atom is m = 3.82×10-26 kg Then (4.80*10-19 C)(212V)=(1/2)(3.82×10-26 kg )v2 Then solving for velocity of the atom we get v =7.3*104m/s Then after entering the magnetic field B=1.25T it traces a circular path Then centripetal force on the atom is provided by force due to magnetic field. Then mv2/r = Bqv Then radius of the circular path r =mv/Bq                                                 =(3.82×10-26 kg (7.3*104m/s) / (1.25T )(4.80*10-19 C)                                                 =4.647*10-3m b) Now sulpur ion is accelerated whose mass is m =5.64×10-26 kg Then work done to accelerate the charge is equal to the change in the kinetic energy of the ion Then q(V)=(1/2)mv2 Then (4.80*10-19 C)(212V)=(1/2)(5.64×10-26 kg )v2 Then v =6*104 m/s Now radius of the circular path is r'= mv/qB                                                    =(5.64×10-26 kg (6*104m/s) / (1.25T )(4.80*10-19 C)                                                    =5.64*10-3m C) The distance between two points in the detector plate is d =r'-r     =5.64*10-3m-4.647*10-3m     =0.993*10-3m Then solving for velocity of the atom we get v =7.3*104m/s Then after entering the magnetic field B=1.25T it traces a circular path Then centripetal force on the atom is provided by force due to magnetic field. Then mv2/r = Bqv Then radius of the circular path r =mv/Bq                                                 =(3.82×10-26 kg (7.3*104m/s) / (1.25T )(4.80*10-19 C)                                                 =4.647*10-3m b) Now sulpur ion is accelerated whose mass is m =5.64×10-26 kg Then work done to accelerate the charge is equal to the change in the kinetic energy of the ion Then q(V)=(1/2)mv2 Then (4.80*10-19 C)(212V)=(1/2)(5.64×10-26 kg )v2 Then v =6*104 m/s Then v =6*104 m/s Now radius of the circular path is r'= mv/qB                                                    =(5.64×10-26 kg (6*104m/s) / (1.25T )(4.80*10-19 C)                                                    =5.64*10-3m C) The distance between two points in the detector plate is d =r'-r     =5.64*10-3m-4.647*10-3m     =0.993*10-3m

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