Question

Electrons in an electron microscope have a kinetic energy of 1.20 105 eV. (a) Find the de Broglie wavelength of the electrons. Incorrect: Your answer is incorrect. (b) Find the ratio of this wavelength to the wavelength of light at the middle of the visible spectrum (550 nm). (c) How many times greater magnification is theoretically possible with this microscope than with a light microscope?

Answer 1

answers::

a)The de Broglie wavelength is given by

  

The electrons here have kinetic energies  

We can use the rest mass for m. So we need to find the speed.

That can be found from the kinetic energy which can, in turn, be found from the potential difference V and the change in potential energy, eV.

So the kinetic energy is

  

and we have potential

  

  

  

This is much shorter than visible light wavelengths (about 500 nm).

b) Visible spectrum wavelength

  

Therefore the ratio of this wavelength to the wavelength of light at the middle of the visible spectrum

Ratio will be = 0.00000636363

Wave effects would be noticed only on a scale much smaller than that at which we notice interference and diffraction effects for light