please answer both questions: question 1: question 1: A doubly ionized molecule

Question

please answer both questions:

question 1:

question 1: A doubly ionized molecule is studied in a mass spectrometer, where it is accelerated through a potential difference of 410 V and found to travel in a circular path of radius 7.70 cm in a 0.460-T magnetic field. What is the molecule’s mass? please give me the answer in kg.

question 2:

question 1:

question 1: A doubly ionized molecule is studied in a mass spectrometer, where it is accelerated through a potential difference of 410 V and found to travel in a circular path of radius 7.70 cm in a 0.460-T magnetic field. What is the molecule’s mass? please give me the answer in kg.

question 2:

question 2: At what speed will a particle of mass 9.7×109 kg and charge 27 nC move in a circular path of radius 5.2 cm in a uniform 2.2-T magnetic field? please give me the answer in m/s. Recording plate AV

Explanation / Answer

Question 1

As particle is accelerated from rest through the potential difference of 410 volt hence final kinetic energy mv2/2 = qV = 2eV. That means mv2 = 4eV = 4*(1.6*10-19)*(410) = 2624*10-19 J.

Now as during its motion in magnetic field

mv2/r = qvB

So

mv2/r = 2evB

v = (2624*10-19)/2erB

= (2624*10-19)/2(1.6*10-19)(0.077)(0.460)

= 23150.76 m/s

So

m = 2624*10-19/(23150.76)2

m = 4.8959*10-25 kg

Question 2

Here as

mv2/r = qvB

v = qBr/m = 0.3184 m/s

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