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.

A parallel electron beam accelerated in an electric field with a poten

1.	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» Solution :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`

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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.

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