Question

You will find in Chapter 39 that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to "observe" such an orbiting electron by using a light microscope to measure the electron's presumed orbital position with a precision of, say, $10pm$ (a typical atom has a radius of about $100pm$ ). The wavelength of the light used in the microscope must then be about $10pm$. (a) What would be the photon energy of this light?(b) How much energy would such a photon impart to an electron in a head-on collision?(c) What do these results tell you about the possibility of "viewing" an atomic electron at two or more points along its presumed orbital path? (Hint: The outer electrons of atoms are bound to the atom by energies of only a few electron-volts.)

Answer 1

(a) The rate at which incident protons arrive at the barrier is

$n=1.0kA/1.60\times {10}^{-19}C=6.25\times {10}^{21}/s$

Letting $nTt=1,$ we find the waiting time $t$

$\begin{array}{c}t=(nT{)}^{-1}=\frac{1}{n}\exp (2L\sqrt{\frac{8{\pi }^2{m}_p(U_b-E)}{h^2}})\\ =\left(\frac{1}{6.25\times {10}^{21}/s}\right)\exp \left(\frac{2\pi \left(0.70nm\right)}{1240eV\cdot nm}\sqrt{8\left(938MeV\right)(6.0eV-5.0eV)}\right)\end{array}$

$=3.37\times {10}^{111}s\approx {10}^{104}y$

which is much longer than the age of the universe.

(b) Replacing the mass of the proton with that of the electron, we obtain the corresponding waiting time for an electron:

$\begin{array}{c}t=(nT{)}^{-1}=\frac{1}{n}\exp [2L\sqrt{\frac{8{\pi }^2{m}_e(U_b-E)}{h^2}}]\\ =\left(\frac{1}{6.25\times {10}^{21}/s}\right)\exp \left[\frac{2\pi \left(0.70nm\right)}{1240eV\cdot nm}\sqrt{8\left(0.511MeV\right)(6.0eV-5.0eV)}\right]\\ =2.1\times {10}^{-19}s\end{array}$

The enormous difference between the two waiting times is the result of the difference between the masses of the two kinds of particles.