# Diffraction on single slit. De Broglie associated wavelength.

1) A parallel beam of monochromatic light passing through an aperture having a narrow rectangular slit placed normal to the direction of the beam, gives diffraction fringes on a screen. Find the the energy and momentum of photons if one knows that the first minimum occurs at an angle $\phi=6 deg$ with the normal to the screen; the slit width is $b = 5 mm$.

The path difference corresponding to the ray emitted by the end of the slit under the angle $\phi$ is

$\delta = b*sin(\phi)$

Since $\delta = (\lambda\phi)/(2\pi)$ we have $b*sin(\phi) =(\lambda\phi)/(2\pi)$.

The first minimum is obtained when the phase difference is $\phi =2\pi$ so that $\lambda = b*sin(\phi)$.

The energy of a photon is

$\epsilon = h\nu=h c/\lambda = (h c)/(b*sin(\phi))= 2.37 meV$

and the corresponding momentum of the photon is

$p =h/\lambda = h/(b*sin(\phi)) =1.27*10^{-30} (kg*m/s)$

2) Somewhere in deep cosmic space an electron approaches a proton. Consider the total energy of the electron at infinitely large distance from proton zero. Find the wavelength $\lambda$ of the wave associated with the electron:

a) at a distance $r=1 m$ from the proton

b) at a distance $r’=0.5*10^{-10} m$ from the proton

c) knowing that $r’$ is of the order of magnitude of the radius of the electron orbit in the ground state of the hydrogen atom compare $\lambda ‘$ with $r’$.

If we take the electron and proton as an isolated system, the total energy is the same at all relative distances:

$E_c + \left ( -\frac{e^2}{4\pi\epsilon_0*r}\right) =0$

so that

$E_c = \frac{e^2}{4\pi\epsilon_0*r}$

The de Broglie wavelength will be

$\lambda = \frac {h}{\sqrt{2mE_c}} = \frac {h}{\sqrt{\frac{2me^2}{4\pi\epsilon_0*r}}} =\frac{h}{e}*\sqrt{\frac{2\pi\epsilon_0*r}{m}}$

With the given distances in text one has:

a) $\lambda = 32.4 \mu m$

b) $\lambda ‘ =2.28*10^{-10} m$

c) $\lambda ‘/r’= 4.56$