An electron is accelerated through a potential difference of 81V. What is the de Broglie wavelength associated with it? To which part of electromagnetic spectrum does this wavelength corresspond ?

Explanation

(i) An electron of mass m is accelerated through a potential difference of V volt. The kineti energy acquired by the electron is given by
`1/2mv^2=eV`
Therefore, the speed v of the electron is `v=sqrt((2eV)/m)`
Hence, the de Broglie wavelength of the electron is `lambda=h/(mv)=h/sqrt(2emV)`
Substituting the known values in the above equation , we get
`lambda=(6.626xx10^(-34))/sqrt(2Vxx1.6xx10^(-19)xx9.11xx10^(-31))`
`lambda=(12.27xx10^(-10))/sqrtV` meter (or) `lambda=12.27/sqrtV` Å
For example, if an electron is accelerated through a potential difference of 100V, then its de Broglie wavelength is 1.227 A. Since the kinetic energy of the electron, K=eV, then the de Broglie wavelength associated with electron can be also written as
`lambda=h/sqrt(2mK)` .
(ii) de-Broglie wavelength of an electron beam accelerated through a potential difference of V volts is
`lambda=h/sqrt(2meV)=1.23/sqrtV` nm
V=81 V , so `lambda=1.23/sqrt81xx10^(-9)m =0.1366xx10^(-9)` m
`lambda`=1.36 Å
X-ray is the part of electromagnetic spectrum does this wavelength corresponds. X-ray has the wavelengths ranging from about `10^(-8)` to `10^(-12)` m .