What is the (a) momentum (b) speed and (c) de-Broglie wavelength of an electron with kinetic energy of 120 eV. Given `h=6.6xx10^(-34)Js, m_(e)=9xx10^(-31)kg , 1eV=1.6xx10^(-19)J`.

Explanation

Given, kinetic energy of electron ,
`E_k=120 eV=120xx1.6xx10^(-19)J=192xx10^(-19)J`
(a) Momentum can also be expressed in terms of kinetic energy as follows :
`impliesp=sqrt(2mE_k)`
`=sqrt(2xx9.1xx10^(-31)xx192xx10^(-19))`
`=5.91xx10^(-24) kg ms"^(-1)`
(b) Speed of an electron is
`v=p/m`
`impliesv=(5.91xx10^(-24))/(9.1xx10^(-31))`
`=6.5xx10^6ms^(-1)`
(c) Wavelength associated with electron is
`lamda=h/p=(6.62xx10^(-34))/(5.91xx10^(24))`
`=1.12xx10^(-10)m`
`implies lamda= 0.112 nm.`