Question

Question #785ad

Answer 1

Rubidium.

Explanation:

The photoelectric effect basically refers to the fact that you can remove an electron from the surface of a metal if you hit that electron with a photon that carries enough energy to overcome the metal's work function.

A metal's work function refers to the amount of energy you need to provide to an electron located on the surface of the metal to get it to a point just outside the surface.

Mathematically, this is written like this

`color(blue)(K_E = h * nu - phi)" "`, where

`K_E` - the kinetic energy of the removed electron;
`h` - Planck's constant, equal to `6.626 * 10^(-34)"J" * "s"`;
`nu` - the frequency of the incoming photon;
`phi` - the work force of the metal.

What this equation tells you is that you can only remove an electron from the surface of the metal if the energy of the incoming photon, `h * nu`, is bigger than the work force.

This will allow the kinetic energy of the removed electron to be positive.

So, the energy of a photon is directly proportional to its frequency, `nu`. This means that its inversely proportional to its wavelength.

Simply put, photons that have longer wavelengths will have lower frequencies, and thus a lower energy. Photons that have shorter wavelengths will have higher frequencies and thus a higher energy.

So what you need to look for is the metal that has the smallest work function, because that will correspond to photons of lower energy, i.e. lower frequency, higher wavelength.

You can check the values of the work function for these metals here

http://public.wsu.edu/~pchemlab/documents/Work-functionvalues.pdf

The smallest value for the work function belongs to rubidium, `"Rb"`. This means that the incoming photons with the smallest frequency, i.e. longest wavelength, will be able to remove electrons from rubidium's surface.

For the other metals in the list you need photons with higher frequency, i.e. shorter wavelength.