Re: Dual nature of light

On Mar 1, 8:14 pm, "Peter" <ptr...@xxxxxxxxx> wrote:
Hello,

I am trying to understand the following statement on photoelectric
effect:

"When light shines on a metal, electrons are emitted. However,
electrons are emitted only at specifc threshold frequency."

I am guess the experimental setup is something as follows:

1. Shine light on a metal.
2. Start at a specific frequency of light.
3. Keep on increasing the frequency of light.
4. At discrete frequencies, you will see (not by naked eyes) eletrons
being emitted.

I would appreciate it if someone can help me understand some
fundamentals:

1. How do you go about changing the frequency of light?
2. Who is losing the electrons? The metal or the light?
3. I guess the atom gets decomposed into electrons and protons. What
happens to the protons?
4. In theory, if light continues to shine on a metal at a specifc
frequency, in a few billion years, the metal will disappear as it
continues to lose atoms. Is this correct?
5. Finally, just because electrons are being emitted at specific
frequencies, how do you conclude that light has discrete particles of
energy?

Thank you in advance for your help.

Pete

No, this isn't quite right. An introductory physics or chemistry text
will describe this in some detail.

Here is the fundamental problem.
- The wave picture of light says that energy gets deposited in a
continuous manner. This means that, even if the frequency is low, if
you wait long enough, enough energy will be deposited to free
electrons. It also means that turning up the *intensity* of the the
light, the time you have to wait will be shorter. It also means that
the electrons will be ejected as soon as they have the chance and
should come off with essentially little or no kinetic energy, or at
least that should be independent of the frequency.
- But that's not what happens in real life. What happens is that if
the frequency is low, it doesn't matter how long you wait or how high
you make the intensity, electrons are *never* ejected. Furthermore, if
you up the frequency (by choosing a different light source, say), even
if the intensity is *low*, you find there is no time delay before
electrons get ejected. If you increase the intensity, there still is
no time delay, it just causes more electrons to be ejected, not
ejected sooner. Finally, the amount of energy the electrons have has a
linear relationship with the light's frequency.
- What happens in real life makes absolutely no sense for a wave
picture of light, which is a real problem, because light certainly
*does* behave like a wave in a lot of other experiments (like Young's
double slit experiment) -- but not in this one.

What Einstein suggested is that the energy is NOT delivered
continuously. It is not delivered any faster if the intensity
increases. It comes in lumps (quanta) that are the size of the
frequency of the light (that is, the energy and the frequency are
proportional), and that energy is deposited all at once for each lump,
take it or leave it. Thus, if you have a lot of lumps that are too
small (too low a frequency) then there is not enough energy delivered
in each lump to kick out an electron, and so none ever appear, no
matter how many there are. Once you cross a certain threshold of
energy (frequency) though, enough energy is delivered right away, and
an electron is kicked out without delay. If the frequency is increased
more, then all of that energy is given to the electron and it carries
away more energy when it goes.

Typically, the experiment involves a closed circuit, with a cathode
and an anode, and when the electrons are ejected, they are picked up
in the other electrode and the electrical circuit (after sensing the
current) delivers the electron to the metal again. So no, the metal
doesn't have to disintegrate.

PD

.

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