By accurately measuring the index of refraction of light from a laser we determine that the wavelength of the light is 760 nm. How much energy is carried by each photon of this light? This light is shined on a photoelectric metal with work function 1.56 electron volts. A metal grid is placed near the metal and held at a variable negative potential with respect to the grid. What is the minimum magnitude of this potential which will prevent any ejected electrons from reaching the grid?
SolutionThe energy of a photon is h * f, where f is the frequency of the electromagnetic wave and h is Planck's constant 6.62 * 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
and photon energy
An electron volt is the energy gained by an electron as it travels through a potential difference of 1 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:
Expressing the photon energy in units of electron volts we have
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 1.56 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 760 nm gives up its total 1.637 of energy to an electron, and when the electron loses no more than the required 1.56 eV, the electron will escape with KE
An electron with this KE can overcome a potential difference of