A photoelectric metal has work function 1.23 electron volts. If it is exposed to monochromatic light with wavelength 460 nm, what is the magnitude of the maximum potential difference through which an ejected electron might travel?

The energy of a photon is h * f, where f is the frequency of the electromagnetic wave and h is Planck's constant 6.63 * 10^-34 J s.

The frequency of the wave is f = c / `lambda, where c = 3 * 10^8 m/s is the speed of light in a vacuum.

We therefore have

• frequency of light: f = 3 * 10^8 m/s / 460 nm = 3 * 10^8 m/s / ( 460 * 10^-9 m) = .6521 * 10^15 Hz

and photon energy

• photon energy = h * f = 6.63 * 10^-34 J s * .6521 * 10^15 Hz = 4.329 * 10^-19 Joules.

An electron volt is the energy gained by an electron as it travels through a potential difference of 0 volt; this energy is equal to the product of the charge 1.6 * 10^-19 C of an electron and the 1 volt = 1 J/C potential difference, or 1.6 * 10^-19 J:

• 1 eV = 1.6 * 10^-19 J.

Expressing this energy in units of electron volts we have

• photon energy = 4.329 * 10^-19 J / (1.6 * 10^-19 J / eV) = 2.705.

When a conduction electron is ejected by a photon it can acquire the entire energy of the photon in the form of KE. However, to escape the metal surface requires a certain energy. This energy is called the work function, equal to 1.23 eV in the present example. It is possible for even more energy to be lost, and it is possible for a photon to give up less than its total energy to an electron, so not all electrons will escape the metal, and not all that escape will have the maximum possible KE as they leave the surface.

When a photon of wavelength 460 nm gives up its total 2.705 of energy to an electron, and when the electron loses no more than the required 1.23 eV, the electron will escape with KE

• escape KE = 2.705 - 1.23 eV = 1.475 eV.

An electron with this KE can overcome a potential difference of

• max PE overcome = 1.475 volts.