J. J. Thomson is best known for his discoveries about the nature of cathode rays.

Question

Mass Spectrometer

J. J. Thomson is best known for his discoveries about the nature of cathode rays. Another important contribution of his was the invention, together with one of his students, of the mass spectrometer. The ratio of mass m to (positive) charge q of an ion may be accurately determined in a mass spectrometer. In essence, the spectrometer consists of two regions: one that accelerates the ion through a potential difference V and a second that measures its radius of curvature in a perpendicular magnetic field. (Figure 1)

The ion begins at potential V and is accelerated toward zero potential. When the particle exits the region with the electric field it will have obtained a speed u.

Part A

With what speed u does the ion exit the acceleration region?

Find the speed in terms of m, q, V, and any constants.

u =

Part B

After being accelerated, the particle enters a uniform magnetic field of strength B0 and travels in a circle of radius R (determined by observing where it hits on a screen--as shown in the figure). The results of this experiment allow one to findm/q in terms of the experimentally measured quantities such as the particle radius, the magnetic field, and the applied voltage.

What is m/q?

Express m/q in terms of B0,V, R, and any necessary constants.

m/q =

[su_box title="♦ Relevant knowledge" box_color="#2d6ff7" radius="1"]Thomson carried out experiments using Thomson conducted experiments with the Cathode Ray Tube which resulted in the discovery of electrons. Thomson's research also revealed characteristics of electrons like being negatively charged along with the attraction of positive charges and also its mass-to-charge ratio. Thomson utilized the Cathode-Ray Tube for his experiments. Thompson's experiments led to the development of a new view of atoms modeling using the equations of law of physics to illustrate the motion of electrons with much accuracy and precision.

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Abigail Anderson · Accepted Answer

It is a given that

Mass of the particle [katex]=\mathrm{m}[/katex]

Mag field [katex]=\mathrm{B} 0[/katex]

Part (A). At what speed does the ion leave the acceleration region?

Let E1 be the initial energy of the particle, which is actually electirc energy. It can also be given by

[katex]E 1=q V[/katex]

The particle's energy is converted to electricity when it moves. [katex]\mathrm{KE}[/katex]

[katex]\mathrm{E} 2=1 / 2 \mathrm{~m} \mathrm{u}^{2}[/katex]

We can save energy by conserving it.

[katex]q V=1 / 2 m u^{2}[/katex]

[katex]\mathrm{u}=\sqrt{(2 q V) / m}[/katex]

[katex]\operatorname{Part}(B)[/katex]

The cyclotron frequency is the (angular frequency) of orbital motion of the electron in the magnetic field. It is given by

[katex]\omega=q \mathrm{BO} / \mathrm{m}[/katex]

We know this by the definition og angular velocity.

[katex]\mathrm{u}=\mathrm{R} \omega \mathrm{So} \omega=\mathrm{u} / \mathrm{R}[/katex]

The two different expressions of [katex]\omega[/katex]

q [katex]\mathrm{BO} / \mathrm{m}=\mathrm{u} / \mathrm{R}[/katex]

Value of [katex]\mathrm{u}[/katex]

q [katex]\mathrm{BO} / \mathrm{m}=\sqrt{(2 q V) / m} / \mathrm{R}[/katex]

We get our sides quarrelled.

[katex]q^{2} B 0^{2} / m^{2}=(2 q V / m) / R^{2}[/katex]

[katex]m / q=R^{2} B 0^{2} /(2 V)[/katex]

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