Could you use a highly intense beam of red light to eject electrons from the surface of a metal?

Answer 1

As a previous poster said, it is not the intensity (i.e., the number of photons per second) impinging on the surface that is of primary importance here; it is the energy of an individual photon.

In order to eject an electron from the surface of a metal, the energy of an incoming photon must exceed the "work function" of the metal. The work function is defined as the minimum energy needed to remove an electron from a solid material to a point just outside the material (see first source). Most metals have work functions between 3 and 6 eV, though some, particularly the alkali metals, are lower (Cs, which has a very low electronnegativity, has a work function of 2.1 eV, for example). See second source for a table of work functions for common metals.

If the photons have sufficient energy to eject electrons, then increasing the intensity of the light will increase the current (the number of electrons ejected per second), but if the photons do not have sufficient energy, increasing the intensity will do nothing.

The energy of a photon is related to its wavelength by the relation:

E = h*c/lambda

where E is the energy of the photon in Joules
h is Planck's constant = 6.6261*10^-34 kg*m^2/s
c is the speed of light = 2.998* 10^8 m/s
lambda is the wavelength in meters

Red light (e.g., the light from a red laser pointer) has a wavelength of about 650 nm = 6.50 * 10^-7 m. Plugging this into the above equation gives:

E = (6.6261*10^-34 kg*m^2/s)*(2.998* 10^8 m/s)/(6.50 * 10^-7 m)
E = 3,056*10^-19 J

1 eV = 1.6022*10^-19 J, so

E = 1.907 eV for 650 nm wavelength light.

(You can also use the applet at the third source to convert between wavelength (or frequency) and energy.)

The energy of 650nm wavelength photon is not high enough to eject an electron from any of the metals listed in the tables in the second source. In order to get a photocurrent (which is what this phenomenon is called) from Cs, with a work function of 2.1 eV, one would need photons with a wavelength of about 591 nm, which corresponds to orange light.

For a more typical metal (e.g. Fe, with a work function of 4.5 eV) one needs photons with a wavelength of ~276 nm, which is in the ultraviolet.

Answer 2

It is possible, but it is highly unlikely. Because Red light has long wavelength, low energy. Some other light with high energy content and low wavelength like Blue light would be preferred more to eject electrons off a metal.

Answer 3

Actually you could use any color light of any intensity (as long as it is above the threshold level of energy necessary to emit one electron). This is called the photoelectric effect, won Albert Einstein the Nobel Prize. But the more intense, and the higher the wavelength of light, the more electrons will be ejected.

Answer 4

No they wouldn't be ejected just excited. The only thing i know that ejects electrons would be X-rays

Answer 5

It is not the number of photons hitting the surface that counts, but the energy of the individual photons.

Answer 6

A ruby laser would probably do it.