Physics, 08.10.2020 14:01, lolmaster101
In the photoelectric effect, light striking the surface of a conductor ejects electrons from the conductor. If we shine light on the anode of a capacitor which has some voltage applied across it, the ejected electrons will be accelerated towards the cathode, and we establish a photocurrent. Using this principle, you would like to build a laser tripwire.
You put a lightbulb (with resistance 10 ), capacitor, and DC voltage source E = 9 V in series, and point a ruby laser of wavelength 694 nm (red light) and intensity I = 1 W/m² at the anode of the capacitor. The anode has workfunction =1.3 eV. Your idea is that as long as the laser is actively shining on the anode, the light bulb will light; if an object obstructs the laser, the light bulb will turn off, and you will notice the light flicker.
A. With the given parameters, with the laser shining on the anode, will current flow in your circuit?
B. One potential problem with your tripwire is that the laser is in the visible spectrum, and a potential burgler might see and avoid it. Will current still flow if you replace the ruby laser with an argon laser, with wavelength 1090 nm (infrared), and double the intensity to 2 W/m? Why or why not?
Answers: 3
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