Atom 1 of mass 35 u and atom 2 of mass 37 u are both singly...

Name: Atom 1 of mass 35 u and atom 2 of mass 37 u are both singly ionized with a charge of +e. After being introduced into a mass spectrometer (see the figure below) and accelerated from rest through a potential difference V = 7.6 kV, each ion follows a circular path in a uniform magnetic field of magnitude B = 0.40 T. What is ?x, the distance between the points where the ions strike the detector?
Uploaded: 2023-03-20T15:54:07-08:00
Duration: 4 min 50 s
Channel: Madhur L
Description: Atom 1 of mass 35 u and atom 2 of mass 37 u are both singly ionized with a charge of +e. After being introduced into a mass spectrometer (see the figure below) and accelerated from rest through a potential difference V = 7.6 kV, each ion follows a circular path in a uniform magnetic field of magnitude B = 0.40 T. What is ?x, the distance between the points where the ions strike the detector?

Atom 1 of mass 35 u and atom 2 of mass 37 u are both singly ionized with a charge of +e. After being introduced into a mass spectrometer (see the figure below) and accelerated from rest through a potential difference V = 7.6 kV, each ion follows a circular path in a uniform magnetic field of magnitude B = 0.40 T. What is ?x, the distance between the points where the ions strike the detector?

Transcript

00:01 Hello students in this question they have given that an autumn of mass 35 unit mass is 35 unit and another atom this is m one another atom of mass 37 unit are both singly ionized with the charge e.

00:28 Q is equal to plus e they have been introduced into the spectrometer you can see in this figure this is how the charge has been introduced in the spectrometer.

00:44 The acceleration from the rest through the potential difference of v is equals to 7 .6 kilo volts.

00:54 Here they have been making a detector which detects the charges which coming towards it.

01:01 So we need to find what is the distance between the points where the ion strike the detector that can be as delta.

01:11 X is equal to question mark.

01:15 We need to find the distance between the points where the ion strike the detector.

01:20 This is the point.

01:24 And they have also given that the magnetic field b is equal to 0 .4 tesla.

01:34 Here if the radius of the curvature, you can see the charge when it is introduced into the spectrometer, it travels in the circular path with the radius r.

01:49 If r is the radius of the curvature, m is the mass of the particle and q is the charge and b is the magnetic field, then we can write r is equals to mv by qb.

02:05 Radius of the path produced by the magnetic field.

02:10 If the velocity v is produced by an accelerating voltage, then we can write v is equals to root of 2qv by where this capital v is the voltage...

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