# Photoelectricity in daily life

Photons strike metals innumerable times in our day-to-day experience.Then, why photoelectrons do not come out of metal surface and cause current?

• Anecdote - I used to frequently receive static shocks from metal screws holding together plastic slides in a certain playground. But only the ones that were exposed to the sun... Jan 26, 2018 at 3:01

The metals that are used in photoelectric experiments belong to the first group of the periodic table. They are often called Alkali metals.

They have the highest electropositive nature in their respective periods. This makes them the most reactive.

Any reactive metal would not exist in nature in its elemental state. So, these metals get oxidised by oxygen present in the air. This makes an oxide layer on their surface. Due to this layer, metals loose their lustre. This layer protects the inner metal from further oxidation or reaction. But if you remove this layer, you can see the lustrous metal again.

So most of the metals that you see aren't really pure. They either exist as oxides or salts. There are unreactive metals like Gold (Work function 5.10$eV$ – 5.47$eV$) which exist in pure metallic state but their work function is quite high compared to highly reactive metals like Caesium (Work function 1.95$eV$).

Metals are reactive due to their tendency to lose electron donors. Therefore, if a metal is unreactive, it is likely to have a high work function.

Light, some infrared, some ultraviolet, and microwaves, pass through the atmosphere and reach the Earth's surface. Gamma rays, X-rays, most of the ultraviolet and some of the infrared are absorbed by the atmosphere and do not reach the Earth's surface. Most of the high energy radiation that is capable of causing photoemission doesn't even reach the Earth's surface.

There can be many other factors actively participating in this case. I think this much information should be enough to answer your question. Cheers!

• So they don't cause current because they no longer react, but if you remove the outer layer completely, they should give current ? Jan 25, 2018 at 10:35
• You need the light to reach the metal, in order to observe photoemission . The oxide layer doesn't let the light pass through. But if you remove the oxide layer and use the metal for the experiment, it will start emitting the electrons again. Jan 25, 2018 at 10:37
• @Mitchell, what happens to light inside the oxide layer then? It can't be completely absorbed there, can it? But reflective properties of the oxide layer (in the optical range) should be worse than the pure metal beneath Jan 26, 2018 at 8:27
• If the oxide layer reflects less, it absorbs more. Non transmitter which is a poor reflector is a good absorber. Jan 26, 2018 at 9:50
• @ChrisH, Then the effect will happen. But not enough to create intense photoemission. Jan 26, 2018 at 12:49

Sometimes the effects do become visible in electronics.

One example is the case where Raspberry Pi 2 could be crashed by camera flashlight:

Upton explained that the semiconductor material used to make the power regulator was subject to a photoelectric effect when hit with light, and if enough light of the right energy was fired at it, then it would "upset" the device, causing it to turn itself off.

• Those were promoted from valence band to conduction band within the silicon, in fact every photodiode on every appliance that has a remote control, every CCD or CMOS camera would then also be an example. edit: oh, I spoke too soon :-)
– uhoh
Jan 26, 2018 at 16:09

They do, but it's too small to notice on a human scale.

On the scale of electronics, you absolutely can see it. We have photoresistors and photodiodes which rely on this effect. You need to be measuring this with a multimeter and looking at changes of resistance though - it's far too small for it to be perceptible as a static shock.

For another use which is now obsolete... Before we had flash memory, we used EPROMs for storing programs on electronic devices. In production you could have a ROM mass-produced (if you were making millions) or use a one-time-programmable PROM, but during development you always needed an EPROM.

The key feature of an EPROM is that it can be electrically programmed, but it is erased by light (specifically UV light). The photoelectric effect causes a build-up of charge everywhere, which knocks all the gates back to their initial state. They can then be electrically programmed, which places a charge on the gates in a specific way. The EPROM chip actually has a transparent window in the centre of the device to expose the gates for erasing. And once you'd programmed an EPROM, you needed to cover the window, otherwise UV from daylight or fluorescent tubes would erase your program.

Yes, photoelectric effect is observed.Photoelectrons come out and form a negatively charged layer above the metal surface.

This prevents more photoelectrons to be emitted.Moreover, to cause the photoelectrons to travel, you require an anode/receiver plate.Ulimately, photoelectric current halts...

• No evidence for such a layer.
– user137289
Jan 25, 2018 at 12:12
• A negatively charged layer above the metal surface? So, in the air? Then what holds it close to the metal, but not so close it recombines? Jan 26, 2018 at 3:10
• I think he is talking about Gauss's law. Jan 27, 2018 at 10:39

A photoelectric current will only arise and be noticed in special circumstances. Heinrich Hertz noticed an effect of UV light on his spark gaps. Hallwachs then showed that light from a carbon-arc lamp would discharge negatively charged gold-leaf electrometers when the light was incident on a zinc plate connected to it. (Zinc has a relatively low work function among surfaces that are stable in air.)

Ordinary indoor light will not cause photoemission from ordinary metal household items. Also, when there is no field, no current would flow even if the photon energy were high enough. The only context where I am aware of possible problems is spacecraft.