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?

a) Ф = 1.79 eV  here are a number of electrons expelled from the metal and these electrons are accelerated by the difference in potential and you can establish a curren

b)  E = 1.14 eV     the energy of the photons is less than the work function of the metal, so there are no ejected electrons and there can be no current

Explanation:

a) This is a problem about the photoelectric effect, for the current to flow there must be ejected electrons so that they can be accelerated, this effect was explained by Insistent using

            h f = K - Ф

where hf is the energy of the incident photons, Ф the metal work function and K the kinetic energy of the ejected electrons, the minimum value that it can have is zero

           Ф = h f

let's use the relationship

           c = λ f

           f = c / λ

we substitute

          Ф = h c / λ

let's calculate

          Ф = 6.63 10⁻³⁴ 3 10⁸/694 10⁻⁹

          Ф = 2.87 10⁻¹⁹ J

let's reduce this value to units of eV

         Ф = 2.87 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

         Ф = 1.79 eV

This is the maximum value that the work function can have for the electrons to be expelled, as they indicate that the work function of the anode is 1.3 eV there are a number of electrons expelled from the metal and these electrons are accelerated by the difference in potential and you can establish a current

B) The laser is changed for another λ = 1090 nm = 1090 10⁻⁹ m

Let's find the energy of the laser photons

         E = h c /λ

         E = 6.63 10⁻³⁴ 3 10⁸/1090 10⁻⁹

         E = 1.82 10⁻¹⁹ J

we reduce to eV

         E = 1.82 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

         E = 1.14 eV

In this case, the energy of the photons is less than the work function of the metal, so there are no ejected electrons and there can be no current

it becomes positively charged by charging it

15.3 cos 65 degrees trig