Before the advent of solid-state electronics, vacuum tubes were widely used in r

Question

Before the advent of solid-state electronics, vacuum tubes were widely used in radios and other devices. A simple type of vacuum tube known as a diode consists essentially of two electrodes within a highly evacuated enclosure. One electrode, the cathode, is maintained at a high temperature and emits electrons from its surface. A potential difference of a few hundred volts is maintained between the cathode and the other electrode, known as the anode, with the anode at the higher potential. Suppose a diode consists of a cylindrical cathode with a radius of 6.20010?2 , mounted coaxially within a cylindrical anode with a radius of 0.5580 . The potential difference between the anode and cathode is 220 . An electron leaves the surface of the cathode with zero initial speed (). Find its speed when it strikes the anode. Express your answer numerically in meters per second. here' some hints: Item 9 Suppose a diode consists of a cylindrical cathode with a radius of 6.20010?2 , mounted coaxially within a cylindrical anode with a radius of 0.5580 . The potential difference between the anode and cathode is 220 . An electron leaves the surface of the cathode with zero initial speed (). Find its speed when it strikes the anode. Hint 1. How to approach the problem Try to draw a simple diagram of the diode, with the path of the electron going from the central cathode to the outer anode. Since the diode is a cylinder, the symmetry implies that only radial motion needs to be considered for the electron, since any motion of the electron around the center will not change its potential energy. Note that only a potential difference is given between the plates, so you will need to be careful about your definitions. Use the equation for the conservation of energy to find the final speed of the electron. Hint 2. Calculate the change in potential energy of the electron Calculate the change in the potential energy of the electron as it moves from the inner cathode to the outer anode. Hint 1. Potential energy and potential Recall that the potential energy of a charge in an electric field is related to the potential of that field, evaluated at the position of the charge, by the equation . In this problem, the potential is not known at either the anode or cathode. However, the potential difference is given, so use the relation above to find the change in the potential energy of the electron. Express your answer numerically in joules. = Calculate the initial kinetic energy of the electron at the cathode. Hint 1. Initial velocity of the electron Recall that the electron is emitted from the surface of the cathode with zero initial speed. Express your answer in joules. =

Explanation / Answer

You have systematically removed all units from the quantities you specify. This makes the question essentially meaningless. However, it is not necessary to know the dimensions of the anode and cathode in order to answer the question, so you are lucky ! However, it is necessary to assume that the voltage is measured in volts. The final KE of the electron is the product q*V where q is the electronic charge 1.6*10^-19C and V is the potential difference through which it moves. This is 350v (I assume). You can equate the energy qV to the usual expression for KE (0.5mv^2) to calculate speed (v). m is the mass of the electron (9.1*10^-31kg) 0.5mv^2 = qV v = sqrt (2qV/m) = 11.1*10^6m/s This is a sufficiently small fraction of the speed of light for relativistic effects to be ignored.

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