In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.

With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.

Part A

What is the most significant conclusion that Thomson was able to draw from his measurements?

He found a different value of q/m for different cathode materials.

He found the same value of q/m for different cathode materials.

From measurements of q/m he was able to calculate the charge of an electron.

From measurements of q/m he was able to calculate the mass of an electron.

Part B

What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.

Express your answer in terms of e, m, d, v0, L, and E0.

Part C

Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.

What is the speed v0 of the electrons in this case?

Express your answer in terms of E0 and B0.

Part D

In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the plates (with no magnetic field present). When you add the magnetic field as described in Part C, to what value do you have to adjust its magnitude B0 to observe no deflection?

Assume that the plates are 6.00 cm long and that the distance between them and the screen is 12.0 cm.

Express your answer numerically in tesla.

Part E

Now suppose you carry out a second Thomson experiment with a different beam that contains two types of particles. In particular, both types have the same mass m as an electron, but one has charge e and the other has charge 2e. The beam is filtered, such that both types of particle have the same speed. As in the previous experiment, initially only the electric field is imposed and the deflection of the beam is observed as a spot on the screen; then, in the second phase of the experiment, one attempts to tune the magnetic field to exactly cancel the effect of the electric field.

What would you observe on the screen during this experiment?

Two off-centered spots in both the first and the second phases of the experiment

Two off-centered spots in the first phase of the experiment; one centered spot in the second phase of the experiment

Two off-centered spots in the first phase of the experiment; one centered and one off-centered spot in the second phase of the experiment

One off-centered spot in the first phase of the experiment; one centered and one off-centered spot in the second phase of the experiment

One off-centered spot in the first phase of the experiment; two off-centered spots in the second phase of the experiment

One off-centered spot in both the first and the second phases of the experiment

A) He finds the same value of q / m for different materials , B) y = ½ (q / m) E L² / v₀ₓ² , C) v = E / B , D) B = 2.13 10⁻⁶ T, E) For the first part I have two off-center points., For the second part I can center one point but the other is off center

Therefore the third statement is correct

Explanation:

Part A

Thomson's experiments are the first proof that the atoms that until now were considered indivisible were constituted by different elements, in these experiments Thomson himself the ratio q / m of several cathodes and always found the same value, which allowed to establish that In atoms there are two types of particles, some of which are mobile and others are still.

When examining his statements the correct one is: He finds the same value of q / m for different materials

Part B

For this part let's use Newton's second law

F = ma

q E = m a

a = q / m E

We use the kinematic relationship

y = voy t - ½ to t²

x = v₀ₓ t

The initial vertical velocity of electrons is zero

y = ½ a (x / v₀ₓ)²

We replace

y = ½ (q / m) E L² / v₀ₓ²

Part C

If there is no deflection, the electric and magnetic forces are the same and in the opposite direction

Fm = Fe

q v B = q E

v = E / B

Part D

We replace

y = ½ (q / m) E L² / (E / B)²

y = ½ (q / m) L² B² / E

If we do not want any deflection the magnetic field has to return the electrons the amount that they lower y = -4.12 cm

-4.12 10⁻² = ½ q / m 0.12² B² / 1.1 10³

-16.97 10⁻⁴ = 6.54 10⁻³ B² q / m

B² = -2.59 10⁻¹ q / m

q / m = -1.758 10¹¹ C/ kg

B = √ 0.259 1.758 10¹¹ = √ 4.55 10⁻¹²

B = 2.13 10⁻⁶ T

Part E

As the charge that the two particles is different

For the first part I have two off-center points.

For the second part I can center one point but the other is off center

Therefore the third statement is correct