Mastering Physics Solutions: An Electron in a Diode

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 (vinitial = 0).

Part A

Find its speed vfinal 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 * 1013
v = 9.16 * 106 m/s

9.16 * 10^6 m/s

Tagged with: