Question

Suppose a diode consists of a cylindrical cathode with a radius of 6.200�10?2cm , mounted coaxially within a cylindrical anode with a radius of 0.5580cm . The potential difference between the anode and cathode is 335V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.

Answer 1

Answer:

Final speed is: $v_f = 5.89\cdot 10^{13}\frac{m}{s}$.

Explanation:

We need to remember the Work-energy theorem which states, in mathematical terms:

$W_{net}=K_f-K_0$

Where $W_{net}$ is the net work done, $K_f$ final kinetic energy  and $K_0$ initial kinetic energy.

Means that the work done will be equal to the change of Kinetic energy.

If the charge of the electron is $e=-1.6\cdot  10^{-19} C$ and the potential difference $V=335V$, then $W_{net}=eV$.

If the mass of the electron is $m=9.1\cdot 10^{-31} kg$,

$K_f=mv_f$

$K_0=mv_i=0$

We have: $W_{net}=K_f-K_0=K_f$ ⇒ $eV=mv_f$

⇒ $v_f = \frac{eV}{m}$

∴ $v_f = 5.89\cdot 10^{13}\frac{m}{s}$