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Phy 122
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Set 7 question 2 states:
Each photon of 462 nm light has energy E = h f, where h is Planck's constant 6.62 * 10^-34 J s.
The frequency of 462 nm light is (3 * 10^8 m/s) / ( 462 nm) = .6493 * 10^15 Hz, so the photon energy is
photon energy = (6.62 * 10^-34 J s) * ( .6493 * 10^15 Hz) = 4.298 * 10^-19 Joules.
At 640 watts / m^2, energy falls on 1 cm^2 at a rate of ( 640 watts / m^2 ) * ( 1 cm^2) = ( 640 watts / m^2) * (10^-2 m)^2 = .064 watts.
The energy falling on this area in a second is
energy in 1 sec = .064 J/s * 1 s = .064 J.
If each photon carries 4.298 * 10^-19 Joules of energy, in one second we will have
number of photons in 1 sec = energy in 1 sec / energy per photon = .064 J / ( 4.298 * 10^-19 J) = .1489 * 10^16.
The average area per photon is therefore 1 cm^2 / ( .1489 * 10^16) = 6.715 * 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 6.715 * 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( 6.715 * 10^-16 cm^2) = 2.591 * 10^-8 cm = 2.591 * 10^-10 m = 2.591 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 6.715 * 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 = 6.715 * 10^-16 cm^2.
Solving for the side s, we obtain
second estimate of distance between photon strikes: dist = s = sqrt [ 6.715 * 10^-16 cm / (4 / `sqrt(3) ) ] = 1.705 * 10^-8 cm = 1.705 Angstroms.
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On a test, would it be necessary to do the step with the rectangle, or can we just use the triangle method and assume that one is more accurate?
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There isn't a lot of difference between the results and I would accept either method, but the triangle does give the better estimate.
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