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?
a. He found a different value of q/m for different cathode materials.
b. He found the same value of q/m for different cathode materials.
c. From measurements of q/m he was able to calculate the charge of an electron.
d. 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

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, 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.

In his experiment, Thomson used a gas sample placed in a region between two charged plates, so a region with a potential difference. In the experiment, a current was observed, suggesting that atoms were broken into charged particles (the electrons, in fact). These electrons, producing this current, were released from the ionisation of the atoms of the gas, and they were originally called cathode rays.

Thomson knew that charged particles could be deflected by magnetic fields, so he used a magnetic field to deflect the cathode rays (the electrons). This way, Thomson was able to measure the ratio of charge to mass, q/m, for these particles. The most relevant result of his experiment was that the ratio q/m for cathode rays was the same for different cathode materials, leading to the conclusion that all cathode rays were consisting of the same particles: the electrons.

So, the correct option is

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

B)

The figure of the problem is missing: find it in attachment.

Here the electron enters from the left, travelling with initial velocity v0, then it is deflected by the presence of the electric field of magnitude E0.

Because of this, the electron is deflected by a certain amount when it reaches the screen. Here, we want to find the difference between the two quantities

[tex]\Delta y_2 - \Delta y_1[/tex]

When the electron is in the region with electric field, its motion is accelerated upward. The vertical acceleration is given by the electric force divided by the mass of the electron:

[tex]a=\frac{F}{m}=\frac{eE_0}{m}[/tex]

Since the vertical motion is an accelerated motion, we can use the following suvat equation to find [tex]\Delta y_1[/tex]:

[tex]\Delta y_1 = ut+\frac{1}{2}at^2 = \frac{1}{2}at^2[/tex]

where u = 0 is the initial vertical velocity, and t is the time taken for the electron to cover the distance d. This time can be found from the horizontal motion, which is uniform:

[tex]t=\frac{d}{v_0}[/tex]

So, we get:

[tex]\Delta y_1 = \frac{1}{2}at^2 = \frac{eE_0}{2m}(\frac{d}{v_0})^2=\frac{eE_0d^2}{2mv_0^2}[/tex]

After exiting the region with electric field, the electron continues its motion with constant velocity. The horizontal velocity is [tex]v_0[/tex], while the vertical velocity is

[tex]v_y = u_y + at = 0+\frac{eE_0}{m}\cdot \frac{d}{v_0} = \frac{eE_0 d}{mv_0}[/tex]

The time taken by the electron to cover the distance L to reach the screen is given by the horizontal motion:

[tex]t'=\frac{L}{v_0}[/tex]

During this time, the vertical distance covered by the electron is:

[tex]\Delta y_2 = v_y t' = (\frac{eE_0 d}{mv_0})(\frac{L}{v_0})=\frac{eE_0 dL}{mv_0^2}[/tex]

Therefore, the difference is:

[tex]\Delta y_2 - \Delta y_1=\frac{eE_0 dL}{mv_0^2}-\frac{eE_0d^2}{2mv_0^2}=\frac{eE_0d}{mv_0^2}(L-\frac{d}{2})[/tex] (1)

C)

In this problem, the total deflection measured is

[tex]\Delta y_2 - \Delta y = 4.12 cm = 0.0412 m[/tex]

Using eq(1), and the following data:

[tex]E_0 = 1.10\cdot 10^3 V/m[/tex] (strength of the electric field)

[tex]e=1.6\cdot 10^{-19}C[/tex] (electron charge)

d = 6.00 cm = 0.06 m is the length of the plates

L = 12.0 cm = 0.12 m is the distance between the plates and the screen

[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electron

We can find v0, the initial velocity of the electron:

[tex]v_0 = \sqrt{\frac{eE_0 d}{m(\Delta y_2 - \Delta y_1)}(L-\frac{d}{2})}=5.03\cdot 10^6 m/s[/tex]

When the magnetic field is activated, the electron will have no deflection only if the electric force and the magnetic force are balanced, so when

[tex]eE_0=ev_0 B_0[/tex]

where [tex]B_0[/tex] is the magnitude of the magnetic field.

Solving for B0, we find:

[tex]B_0 = \frac{E_0}{v_0}=\frac{1.10\cdot 10^3}{5.03\cdot 10^6}=2.2\cdot 10^{-4}T[/tex]