Question

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.

Answer 1

Answer:

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

b)      y = $- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}$ ,  c)  v = $\frac{E_o}{B_o}$

Explanation:

In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.

b) In the part of the plates the electrons are accelerated by the electric field,

              F = ma

             - e E = m a

              a = - (e/m)  E₀

               

the distance traveled is          

X axis

          x = v₀ t

the separation of the plates is x = d

          t = vo / d

               

Y axis

          y = v_{oy} t + ½ to t²

          y = ½ a t²

          y = $- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}$

           

c) In this case there is a magnetic field B₀ and the electrons have no deflection

         F = - e E + e v x B

       

if there is no deviation F = 0

         e E = e v B

         v = $\frac{E_o}{B_o}$