The intensity of 836 nm light falling on a photoelectric metal is 110 watts / m^2 (maximum sunlight intensity at the surface of the Earth is on the order of 1000 watts/m^2). How many photons fall on a 1 cm^2 area in a second? What is the approximate average spacing of the photons that fall on this area in a second?

Each photon of 836 nm light has energy E = h f, where h is Planck's constant 6.62 * 10^-34 J s.

The frequency of 836 nm light is (3 * 10^8 m/s) / ( 836 nm) = .3588 * 10^15 Hz, so the photon energy is

• photon energy = (6.62 * 10^-34 J s) * ( .3588 * 10^15 Hz) = 2.375 * 10^-19 Joules.

At 110 watts / m^2, energy falls on 1 cm^2 at a rate of ( 110 watts / m^2 ) * ( 1 cm^2) = ( 110 watts / m^2) * (10^-2 m)^2 = .011 watts.

The energy falling on this area in a second is

• energy in 1 sec = .011  J/s * 1 s = .011  J.

If each photon carries 2.375 * 10^-19 Joules of energy, in one second we will have

• number of photons in 1 sec = energy in 1 sec / energy per photon = .011   J / ( 2.375 * 10^-19 J) = .04631 * 10^16.

The average area per photon is therefore 1 cm^2 / ( .04631 * 10^16) = 21.59 * 10^-16 cm^2.

To make our first estimate the avearage distance between photons we might assume that the photons each strike at the center of a rectangle whose area is 21.59 * 10^-16 cm^2. The approximate average distance between photons would therefore be close to the length of a side of the rectangle. The length of each side is thus

• first estimate of distance between photon strikes: dist = `sqrt( 21.59 * 10^-16 cm^2) = 4.646 * 10^-8 cm = 4.646 * 10^-10 m = 4.646 Angstroms..

This can be compared with the approximate 1-Angstrom diameter of most atoms.

This estimate is pretty good, but leaves each 'strike' surrounded by four atoms with the calculated spacing, and four more with `sqrt(2) times this spacing (to see this sketch a grid of squares with a dot at the center of each, and look at the 8 dots nearest to a given dot).

A better estimate would assume that each photon strikes at the center of an equilateral triangle whose area is 21.59 * 10^-16 cm^2. In this model the atoms will form a honeycomb pattern, with each 'strike' from its nearest neighboring 'strike' by a distance equal to the side of a triangle. If s is the length of a side the area of each triangle is `sqrt(3) / 4 * s^2 (this is a standard result from simple geometry and can be easily derived by using the Pythagorean Theorem to find the altitude of the triangle to be `sqrt(3) / 2 * s and multiplying 1/2 base * altitude to get area). We thus have

• equilateral triangle model for spacing:  `sqrt(3) / 4 * s^2 = 21.59 * 10^-16 cm^2.

Solving for the side s, we obtain

In general if we have beam intensity I, in watts / m^2, and wavelength `lambda, we can calculate the number of photons striking per unit of area in any time interval `dt.

We first calculate the energy of a photon as

• photon energy: Ephoton = h f = h (c / `lambda).

We then calculate the energy per unit area for the time interval `dt:

• energy per unit area in time `dt: E / A = I * `dt.

The number of photons striking per unit area in time `dt will therefore be the energy per unit area divided by the energy per photon:

• number of photons per unit area in time `dt = ( E / A ) / Ephoton = I `dt / ( h c / `lambda) = I `lambda `dt / (h c),

which is also the same as I `dt / (h f ).

If we assume that the photons strike at equal distances in a 'honeycomb' pattern of equilateral triangles, we find that the separation will be

• separation by triangle model = `sqrt( 1 / [ I `dt / ( h f ) ] = `sqrt( h f / (I `dt) ).