(a) Show that the de-Broglie wavelength `lamda` of electrons of kinetic energy K is given by the relation
`lamda=(h)/(sqrt(2mK))`
(b) Calculate the de-Broglie wavelength of an electron beam accelerated through a potential difference of 60 V.

Explanation

(a) We know that de-Broglie wavelength of an electron `lamda=(h)/(P)`
If K be the kinetic energy and m the mass of electron, then `p=sqrt(2mK)`, and hence relation for de-Broglie wavelength becomes:
`lamda=(h)/(sqrt(2mK))`
(b) Here V=60V and we know that for electron `e=1.6xx10^(-19)C, m=9.11xx10^(-31)kg and h=6.63xx10^(-34)Js`
`therefore`de-Broglie wavelength
`lamda=(h)/(sqrt(2meV))=(6.63xx10^(-34))/(sqrt(2xx9.11xx10^(-31)xx1.6xx10^(-19)xx60))m`
`=1.586xx10^(-10)m or 1.586Å`.