Study Guide 4 - Diffraction, Electromagnetic Waves and Particle Waves

\(\mathrm {The\; relationship\; between\; a, \;r \;and\; q\; is;}\)
\(\mathrm {a = r\; \theta}\)
\(\mathrm {when\; \theta \;  is\; in\; radians.}\)
\(\mathrm {q  = a/r = 2.4/50\times 1000\times 100 = 4.8\times 10^{-7} \;radians}\)

\(\mathrm {a = 1.5/d = 1.22\; \lambda/a = (1.22\times 550\times 10^{-9} )/(2\times 10 ^{(-3)})}\)
\(\mathrm {d = 4.47\times 10^3 \;m}\)

Decreasing wavelength
\(\mathrm {Radio \;waves: \lambda \sim 1 - 10 \;m}\)
\(\mathrm {Infrared: \lambda \sim 10^{-6} \; m}\)
\(\mathrm {Red\; light: \lambda \sim 8000\times 10^{-10} \; m}\)
\(\mathrm {Blue\; light: \lambda \sim 4000\times 10^{-10}\; m}\)
\(\mathrm {Ultraviolet: \lambda \sim 2000\times 10^{-10} \;m}\)
\(\mathrm {X-rays: \lambda \sim 100\times 10^{-10} \; m}\)

(a) \(\mathrm {f = 1.46\times 106 \;Hz}\)

\(\mathrm {c = f\lambda \;where\; c = 3.00\times 108\; m/s}\)
\(\mathrm {Therefore\; \lambda = c/f = 3\times 108/1.46\times 106 = 205 \;m}\)

(b) \(\mathrm {f = 1.46\times 106 \;Hz}\)

\(\mathrm {E = hf \;where\; h \;is \;Planck's\; constant = 6.63\times 10^{-34} \; J.s}\)
\(\mathrm {E = (6.63\times 10^{-34})(1.46\times 10^6) = 9.68\times 10^{-28}\; J}\)

\(\mathrm {E = hc/\lambda}\)

\(\mathrm {In\; this \;case\; (\lambda) = 0.10\times 10^{-9} \; m}\)
\(\mathrm {E = hc/\lambda = (6.63\times 10^{-34})(3.00\times 10^8/0.10\times 10^{-9})}\)
\(\mathrm {= 1.2\times 10^{-15}\; J}\)

\(\mathrm {The\; CJOY\; photon\; from\; Question \;4\; had\; energy\; 9.68\times 10^{-28} \; J.}\)
Number of CJOY photons required to make the energy of one X-ray photon
\(\mathrm {= 1.2\times 10^{-15}/9.68\times 10^{-28} = 2.1\times 10^{12}}\)

\(\mathrm {\lambda = h/mv}\)

\(\mathrm {From\; Question\; 5, \lambda = 0.10\; nm \;and \;knowing \;that \;the \;mass \;of \;the \;neutron \;is \; 1.67\times 10^{-27} \; kg}\)
\(\mathrm {\lambda = h/mv}\)
\(\mathrm {v = h/m\lambda = 6.63\times 10^{-34}/(1.67 \times 10^{-27})(0.10 \times 10^{-9})}\)
\(\mathrm {= 3970 = 4.0 \times 10^3 \;m/s}\)

First of all we need to find the wavelength of the electron. To do this we need its speed. What is the expression for the Kinetic Energy (KE) of a particle?

\(\mathrm {KE = (1/2)mv^2}\)  OR \(\mathrm {KE = mv}\)
 

\(\mathrm {KE = (1/2)mv^2 \;is \;correct}\)

\(\mathrm {Therefore\; v^2 = 2(KE)/m \\ = 2(6.00\times10^{-22})/9.11\times 10^{-31}) = 1.3172 \times 10^9 \\ v = 3.629\times 10^4 \;m/s}\)

\(\mathrm {Now \;using \;deBroglie's \;equation: \\ \lambda = h/mv = 6.63 \times 10^{-34}/(9.11 \times 10^{-31} \times 3.629 \times 10^4) \\ = 2.005 \times 10^{-8} \; m}\)

\(\mathrm {Now\; using\; Planck's\; equation: \\ E = hc/\lambda = 6.63 \times 10^{-34} \times 3 \times 10^8/2.005\times 10^{-8} \\ = 9.92 \times 10^{-18} \; J}\)

OR a more elegant and shorter solution:

\(\mathrm {KE = (1/2)mv^2 \\ v^2 = 2KE/m \\ v = (2KE/m)^{1/2} \\ \lambda = h/mv or v = h/m\lambda \\ Therefore, h/ml = (2KE/m)^{1/2}}\)

\(\mathrm {Multiply\; both\; sides\; by\; mc; \\ hmc/m\lambda = hc/\lambda = (2KE/m)^{1/2}mc \\ E = hc/\lambda = (2Kem)^{1/2}c \\ = [2(6.00 \times 10^{-22})(9.11 \times 10^{-31})]^{1/2}3.0 \times 10^8 \\ = 9.92 \times 10^{-18} \; J\; as\; before.}\)

Notice that the value of Planck's constant h was not needed.
 

What is the equation for the Probability density?

\(\mathrm {\Psi = \sqrt \frac{2}{\ell} \sin \frac {n\pi x}{\ell}}\)  OR  \(\mathrm {\Psi ^ 2 = \frac {2}{\ell} \; \sin ^2 \frac{n\pi x}{\ell}}\)

The probability density is:
\(\mathrm {\Psi ^2 = \frac {2}{\ell} \; \sin ^2 \frac{n\pi x}{\ell}}\)

\(\mathrm {The\; value\; of\; \Psi^2 \; is\; a\; maximum\; when\; \sin n\pi x/\ell \;is\; a\; maximum.\\\;The\; maximum\; of\; any\; sine \;is\; 1, so: \Psi^2\;_{max} = 2/\ell}\)

What is length of the retinal \(\mathrm {\pi-chain,}\) which has 12 atoms?

\(\mathrm {11\times 0.15 = 1.65 \;nm}\)

\(\mathrm {So\; the\; maximum\; value\; of\; \Psi^2\; is:\Psi^2 = 2/\ell = 2/1.65 = 1.21 \;nm^{-1}}\)

 

If n = 3 then there are 3 maxima in the graph of \(\mathrm {\Psi^2.}\) It looks like this:



The maxima are at (1/6)l, (1/2)l and (5/6)l where l = 1.65 nm
 

The graph of \(\mathrm {\Psi^2}\) vs \(\ell \) looks like this:



The probability is zero at 0, \((1/3)\ell\), \((2/3)\ell\) and \(\ell\)

9b where b is the bond length.

Now find the probability density (probability per unit length) at the position of the 10th atom.

\(\mathrm {\Psi^2 = \frac{2}{\ell} \sin^2 \frac{n \pi x}{\ell}}\)

\(\mathrm {Where\; \ell = 11b, n = 3 \; and \; x = 9b}\)

\(\mathrm {\Psi^2 = [2/(11 \times 0.15)] \sin^2[3\pi (9b)/(11b)] \\ = 1.21 \sin^2(27/11) \pi \; {angle\; in \; radians} \\ = 1.21 \sin^2(27/11)180^\circ \;{angle \; in \; degrees} \\ = 1.21 \sin^2 442^\circ \\ = 1.21 \times (0.99)^2 = 1.19 \; nm^{-1}} \\\)

Probability of finding the electron in a distance 0.0050 nm centered at the 10th atom
\(\mathrm {= 1.19\; nm^{-1} \times 0.0050 \;nm = 5.9\times 10^{-3}}\)

Look at the \(\Psi^2\) graph:

There is clearly 1½ wavelengths in the length of the chain which is 1.65 nm long. Therefore the electron wavelength is (2/3) 1.65 = 1.10 nm.

The probability distribution for this chain is shown below with the position of the 10 atoms indicated.

The probability of finding the electron somewhere on the chain is 1. This is the total area under the graph.
Clearly the probability of finding the electron somewhere between atoms 1 and 7 is 2/3 of this i.e., \(\mathrm {(2/3)\times1 = 2/3.}\)