A parallel electron beam accelerated in an electric field with a potential difference of 15 V falls on a narrow rectangular diaphragm 0.08 mm wide. Find the width of the principal diffraction maximum on a screen placed 60 cm away from the diaphrugm.

Answer

`4.8mum`

Explanation

The de Broglie wavelength for these electrons is `lamda=h//sqrt(2mevarphi)=12.25//sqrt15=3.2Å`. The first-order diffraction minimum is observed at an angle `theta` such that `sintheta=lamda//D`, where p is the width of the slit (57.9). Since the angle is very small the width of the principal maximum is
`x=2l tantheta=2llamda//D`