**An Electron in a Diode**

Part A = 0.466 m

**Solution Below:**

Suppose a diode consists of a cylindrical cathode with a radius of 6.200*10

^{-2}cm, mounted coaxially within a cylindrical anode with a radius of 0.5580 cm. The potential difference between the anode and cathode is 260 V. An electron leaves the surface of the cathode with zero initial speed (v_{initial}= 0).

Part A

Find its speed v

Express your answer numerically in meters per second.

_{final}when it strikes the anode.Express your answer numerically in meters per second.

The size of the cathode and anode can be ignored – the electric potential is what matters here. Start by figuring out the energy difference due to the electric potential. This is equivalent to kinetic energy:

KE = qV

KE = 1.602 * 10^{-19} * 260

KE = 4.165 * 10^{-17}

Since KE = 1/2mv^2, set this formula equal to the above (note that the charge on an electron and the mass of an electron are just constants that you should memorize):

mv^2 = 8.330 * 10^{-17}

9.11 * 10^{-31} * v^2 = 8.330 * 10^{-17}

v^2 = 9.14 * 10^{13}

v = 9.16 * 10^{6} m/s

9.16 * 10^6 m/s