what is the most significant conclusion that thomson was able to draw from his measurements?

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

B) $\frac{eE_0d}{mv_0^2}(L-\frac{d}{2})$

C) $2.2\cdot 10^{-4}T$

Explanation:

J. J. Thomson was an experiment realized for the get-go time in 1897 that lead to the discovery of the electron.

In his experiment, Thomson used a gas sample placed in a region between two charged plates, then 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 about 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 aforementioned value of q/m for unlike cathode materials.

The figure of the problem is missing: notice 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. Hither, we desire to notice the difference between the ii quantities

$\Delta y_2 - \Delta y_1$

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

$a=\frac{F}{m}=\frac{eE_0}{m}$

Since the vertical motion is an accelerated motion, nosotros tin utilize the following suvat equation to find $\Delta y_1$ :

$\Delta y_1 = ut+\frac{1}{2}at^2 = \frac{1}{2}at^2$

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

$t=\frac{d}{v_0}$

So, nosotros go:

$\Delta y_1 = \frac{1}{2}at^2 = \frac{eE_0}{2m}(\frac{d}{v_0})^2=\frac{eE_0d^2}{2mv_0^2}$

Afterward exiting the region with electrical field, the electron continues its motion with constant velocity. The horizontal velocity is $v_0$ , while the vertical velocity is

$v_y = u_y + at = 0+\frac{eE_0}{m}\cdot \frac{d}{v_0} = \frac{eE_0 d}{mv_0}$

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

$t'=\frac{L}{v_0}$

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

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

Therefore, the deviation is:

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

In this problem, the total deflection measured is

$\Delta y_2 - \Delta y = 4.12 cm = 0.0412 m$

Using eq(1), and the post-obit data:

$E_0 = 1.10\cdot 10^3 V/m$ (strength of the electrical field)

$e=1.6\cdot 10^{-19}C$ (electron charge)

d = half dozen.00 cm = 0.06 chiliad is the length of the plates

50 = 12.0 cm = 0.12 g is the altitude between the plates and the screen

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

Nosotros tin find v0, the initial velocity of the electron:

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

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

$eE_0=ev_0 B_0$