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

Atom 1 of mass 35 $\mathrm{u}$ and atom 2 of mass 37 $\mathrm{u}$ are both singly ionized with a charge of $t$ . After being introduced into a mass spectrometer $($ Fig. $28-12)$ and accelerated from rest through a potential difference $V=7.3 \mathrm{kV},$ each ion follows a circular path in a uniform magnetic field of magnitude $B=0.50 \mathrm{T}$ . What is the distance $\Delta x$ between the points where the ions strike the detector?

Transcript

0:00 Hi there.

00:01 So for this problem, we have an atom that we are going to call of mass m1 that is equal to 35 units of mass and an atom 2 of mass m2 that it has 37 units of mass and are both singly ionized with a charge of t.

00:23 So after being introduced into a mass spectrometer and accelerated from rubeau, rest through epitomical difference that is equal to 7 .3 kilo -boles.

00:42 And each ion follows a circular path in an uniform magnetic field with a magnitude that is equal to 0 .50 tesla.

00:59 And so for this problem, we are asked, what is the distance delta adds between the points where the ions strike the detector? so to calculate that, that is kind of the uncertainty for delta x, we start with the expression for the mass in this circular path.

01:25 We know that the mass is equal to the magnetic field to the squared times the charge, times the distance adds to the square divided by a times the potential difference.

01:41 So in here, we need to first differentiate this with respect to x.

01:47 So we're going to have that delta m is going to be equal to the magnetic field to the square times the charts, divided by 8 times the both, the potential difference.

01:57 And this times this derivative in here that we are going to obtain that that is 2 times x times delta x.

02:05 So we put that in here.

02:10 And now what we are going to do is to also solve from here to x.

02:18 So we're going to have that x from that equation is equal to the square root of a times the potential difference times the mass divided by the magnetic field to the squared times the charge.

02:36 So in here we introduced this into the previous expression, and by doing that we obtain that delta in the mass, the uncertainty on the mass is going to be equal to, after we substitute, we're going to obtain that this is the magnetic field times the square root of the mass times the charge divided by two times the potential difference, and this times the uncertainty on delta, on the distance x.

03:16 Does the distance between the spots made on the photographic plate, we just solved from this equation, and we will find that delta x is simply equal to the change in the mass, divided by the magnetic field times the square root of two times the potential difference, divided by the mass times the charge.

03:46 And now we substitute all of these values...

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