- #1

- 7

- 0

If anyone could help me it would be great :)

Thanks a lot

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter andreitta
- Start date

- #1

- 7

- 0

If anyone could help me it would be great :)

Thanks a lot

- #2

- 26

- 0

1. The electric current $I$ at the collector is (micro mechanism):

I = n s e v

Here, $n$ is a constant relatively (depends on the material and temperature of the negative emitter), $e$ the constant of charge of an electron, $S$ a constant relatively (the cross section of the tube). Hence

I = I(v)

where $v$ is the independent variable, which is determined by the accelerating voltage in the tube, and the small decelerating voltage(U_2) near the destination collector.

2. The magnitudes of velocity of the electrons initially emitted are not the same, the electrons' emitting speeds form a Maxwell distribution, with one peak (most probably values). Some has the very speed (energy-eV) to be absorbed, while some not.

Now, let's turn to the $U_1$ vs $I_a$ graph.

1. Between a valley (on the left with smaller U_1) and a nearest peak on its right, the slope goes upwards and $I_a$ increases.

Because: Most electrons' speeds (energy) are improper to be absorbed. They keep accelerating. The larger $v$, the larger $I_a$.

2. As to the valleys, where most electrons' speeds are proper and therefore their energy are absorbed, and they could no longer go through the decelerating voltage(U_2) to arrive the collector. However, the minor proportion of the electrons donot suffer from the absorbing of kinematic energy, and could cross the decelerating voltage(U_2) to generate $I_a$ in the destination collector. At the valleys, it is this minor electrons, whose motion are not perturbed, that make a difference. Suppose that the proportion is relatively constant (for a not so long interval of U_1 where there are still several valleys and peaks), then $n(minor)$ is constant, then $I_a(valley)$ depends on $v(valley)$. Just as above, the larger $U_1$, the larger $I_a$ at the valleys. So, minimums increase, too.

Share: