Question

An electron is accelerated through a potential difference of 64 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?

Answer 1

de -Broglie wavelength,`lambda = h/(sqrt(2meV))`

Where,

`m ->\text { Mass of electron} = 9.1 xx 10^-31kg`

`lambda = (6.626 xx 10^-34)/(sqrt (2 xx 9.1 xx 10^-31 xx 1.6 xx 10^-19  xx 64))`

`lambda = (6.626  xx 10^-34)/(sqrt 1863.63, xx 10^-50)`

`lambda = (6.626  xx 10^-34)/(sqrt 1063.68 xx 10^-25)`

`=6.626/43.17  xx 10^-9`

`=0.15 xx 10^-9 m`

` =1.5 xx 10^-10 m`

This value of wavelength corresponds to the X-ray region of the electromagnetic spectrum.