The kinetic energy (K.E.) of a beam of electrons, accelerated through a potential V, equalws the energy of a photon of wavelength of 5460 nm. Find the de-Broglie wavelength associated with this beam of electrons.

Explanation

`E=hv=(hc)/(lambda)=e`
`V=(hc)/(lambda e) =(6.6xx 10^-34xx 3xx10^8)/(5460xx 10^-9-1.6xx10 ^-19)=0.22V`
Using , ` lambda =(12.3)/(sqrt(V))A^@rArr lambda=(12.3)/(sqrt(0.22)) =26.2Å`.