(II) Suppose the electric field between the electric plates...

(II) Suppose the electric field between the electric plates in the mass spectrometer of Fig. 33 is $2.48 \times 10^{4} \mathrm{V} / \mathrm{m}$ and the magnetic fields are $B=B^{\prime}=0.58 \mathrm{T}$ . The source contains carbon isotopes of mass numbers $12,13,$ and 14 from a long dead piece of a tree. (To estimate atomic masses, multiply by $1.66 \times 10^{-27} \mathrm{kg} .$ ) How far apart are the lines formed by the singly charged ions of each type on the photographic film? What if the ions were doubly charged?

(II) Suppose the electric field between the electric plates in
the mass spectrometer of Fig. 33 is $2.48 \times 10^{4} \mathrm{V} / \mathrm{m}$ and the
magnetic fields are $B=B^{\prime}=0.58 \mathrm{T}$ . The source contains
carbon isotopes of mass numbers $12,13,$ and 14 from a long dead piece of a tree. (To estimate atomic masses, multiply by
$1.66 \times 10^{-27} \mathrm{kg} .$ ) How far apart are the lines formed by the
singly charged ions of each type on the photographic film?
What if the ions were doubly charged?

Key Concepts

Ion Charge States

Ion charge states refer to the net electrical charge on an ion, typically resulting from the loss or gain of electrons. In mass spectrometry, whether an ion is singly, doubly, or multiply charged directly affects its trajectory in a magnetic field, as the radius of curvature is inversely proportional to the charge (r=mv/(qB)). Thus, changes in the ion's charge state lead to different deflection paths, which are key to resolving ions of similar mass but different charges.

Mass-to-Charge Ratio (m/q)

The mass-to-charge ratio is a critical parameter in mass spectrometry that determines how an ion will respond to electric and magnetic fields. Ions with different masses or different levels of ionization (charge states) will have different m/q values, leading to distinct trajectories when subjected to electromagnetic forces. This difference is what allows a mass spectrometer to distinguish between isotopes and molecular species.

Lorentz Force

The Lorentz force is the force exerted on a charged particle moving through electric and magnetic fields, defined by the equation F=q(E+v×B). In the context of a mass spectrometer, while the electric field is responsible for accelerating the ions, the magnetic component of the Lorentz force provides the centripetal force that bends the ions' paths. The resulting deflection is used to separate ions based on their mass-to-charge ratio.

Electric Field Acceleration

In devices like a mass spectrometer, an electric field is used to accelerate ions from a source to a specific velocity. The electric force acting on a charged particle (F=qE) imparts kinetic energy to the ion, with its velocity determined by both the magnitude of the electric field and the charge of the ion. This acceleration step is critical for establishing the conditions required for subsequent deflection in a magnetic field.

Mass Spectrometry

Mass spectrometry is an analytical technique used to separate, identify, and quantify ions based on their mass-to-charge ratio (m/q). In a mass spectrometer, ions are accelerated and then passed through regions of electric and magnetic fields, which cause them to follow different trajectories based on their m/q ratio. This separation allows for the detection and identification of distinct isotopes or molecular fragments.

Magnetic Field and Charged Particle Deflection

Once the ions are accelerated, they enter a region with a magnetic field where they experience a Lorentz force perpendicular to both their velocity and the magnetic field (F=qvB). This force results in the ions moving along curved paths. The radius of curvature of these paths is given by r=mv/(qB), linking the particle's mass and charge with its motion in the magnetic field.

Transcript

00:01 Okay, so during chapter 27, property 53, says suppose the electric field between the electric plates and the mass spectrometer of figure 23 is 2 .48 times 10 to the 4 volts per meter.

00:16 So this is our electric field magnitude.

00:19 And the magnetic fields are b equals b prime equals 0 .58 tesla.

00:26 The source contains carbon isotopes of mass numbers 12.

00:33 So i'm going to write carbon of mass numbers 12, 13, and 14 from a long dead piece of tree to estimate atomic masses multiplied by it.

00:46 Okay.

00:47 How far apart are the lines formed by the singly charged ions of each type on the photographic film? what if the ions were doubly charged? so the location of each line on the film is just equal to twice the radius of curvature.

01:04 And the radius of curvature can be found from the expression given in section 27 line, if you want to look that up.

01:12 But we take m, this is equal to qb, b prime, r over e.

01:19 So if we rearrange for r, this is m, e, over q, b, prime.

01:29 So if we're wanting to find 2r, which is the value we want to find, this equals what? so 2r equals 2mb over q.

01:41 And for this problem, we know b equals b prime, so we just put b squared.

01:48 Okay.

01:49 So if we just find this value for each of our different masses, that's all we got to do.

01:58 We know that 2r for 12 is given by 2 times 12 times 1 .67 times 10 to the meter 27 kilograms to convert to kilograms times e 2 .48 times 10 to the 4 volts per meter.

02:19 And we can divide this by q, 1 .6 times 10 to the negative 19 coulums, and our b, 0 .58 squared.

02:32 So this comes out to mean 1 .8467 times 10 to the negative 2 meters.

02:44 Okay, so we can now find 2 r13, and this is the same thing.

02:51 We just multiply by 13 now, instead of 12...

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