Explanation:
Like all EM waves, light transports energy across space. The intensity (energy per unit area and unit time) is proportional to the square of the amplitude of the electric field of the light wave. This energy, however, arrives at a receiver not continuously but in discrete units called photons. The energy transported by an electromagnetic wave is not continuously distributed over the wave front. It is transported in discrete packages. Photons are the particles of light.
Properties of photons:
Photons always move with the speed of light.
Photons are electrically neutral.
Photons have no mass, but they have energy E = hf = hc/λ. Here h = 6.626*10-34 Js is a universal constant called Planck's constant. The energy of each photon is inversely proportional to the wavelength of the associated EM wave. The shorter the wavelength, the more energetic is the photon, the longer the wavelength, the less energetic is the photon.
A laser beam and a microwave beam can carry the same amount of energy. In this case the laser beam contains a smaller number of photons, but each photon in the laser beam has a higher energy than the photons in the microwave beam.
Photons can be created and destroyed. When a source emits EM waves, photons are created. When photons encounter matter, they may be absorbed and transfer their energy to the atoms and molecules. Creation and destruction of photons must conserve energy and momentum. The magnitude of the momentum of a photon is p = hf/c = h/λ.
Problem:
What is the energy of a photon of blue light (λ = 450 nm) and of a photon of red light (λ = 700 nm) in units of eV = 1.6*10-19 J?
Solution:
E = hc/λ.
Blue light: E = (6.626*10-34 Js)(3*108 m/s)/(450*10-9 m) = 4.4*10-19 J = 2.76 eV
Red light: E = (6.626*10-34 Js)(3*108 m/s)/(700*10-9 m) = 2.8*10-19 J = 1.8 eV
Problem:
A 100-W incandescent light bulb uses 100 W (100 W = 100 J/s) of electrical power but only radiates about 15 W of actual visible light. Roughly how many visible photons per second hit the open pages of a typical hardcover book if the pages are about 2 m from the bulb and face it directly?
Solution:
We have to make a variety of estimations and approximations to solve this problem.
To find the number of photons hitting the pages each second, we have to know the light energy hitting the pages per second and the energy per photon. We could compute the latter if we knew the wavelength of the light, but the visible light emitted by a normal incandescent bulb is a mix of wavelengths.
Let us estimate that the average wavelength of the visible light is about 550 nm, which is in the yellow region of the spectrum. This means that the average energy per photon is about
E = hc/λ = (6.626*10-34 Js)(3*108 m/s)/(550*10-9 m) = 3.61*10-19 J = 2.26 eV.
To find the number of photons hitting the pages of a book, we need to know the energy per second that falls on the pages. Let us assume that the light energy from the bulb travels uniformly in all directions. Imagine a sphere centered on the light bulb with a radius R = 2 m. Every second, 15 J or so of visible light energy crosses this sphere. If the light energy is spread uniformly over the sphere, then the intensity of the visible light or the energy per unit area per unit time in units of W/m2 at any point on the inner surface of the sphere will be
I = Plamp/(4πR2),
Electrons are ejected from a metal surface with speeds ranging up to 4.6*105 m/s when light with a wavelength of 625 nm is used.
(a) What is the work function of the surface?
(b) What is the cutoff frequency of the surface?
