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

Kinetic energy of the electron , `E_k=120eV`
Planck's constant, `h=6.6xx10^(-34) Js`
Mass of an electron , `m=9.1xx10^(-31) kg`
Charge on an electron , `e=1.6xx10^(-19) C`
(a)For the electron , we can write the relation for kinetic energy as : `E_k=1/2 mv^2`
where,
v=Speed of the electron
`therefore v^2=sqrt((2eE_k)/(m))`
`=sqrt((2xx1.6xx10^(-19)xx120)/(9.1xx10^(-31)))`
`=sqrt(42.198xx10^(12)) = 6.496xx10^6` m/s
Momentum of the electron, p=mv
`=9.1xx10^(-31)xx6.496xx10^6`
`=5.91xx10^(-24) kg ms^(-1)`
Therefore, the momentum of the electron is `5.91xx10^(-24) kg ms^(-1)`
(b)Speed of the electron, `v=6.496xx10^6` m/s
(c )De Broglie wavelength of an electron having a momentum p, is given as : `lambda=h/p`
`=(6.6xx10^(-34))/(5.91xx10^(-24))=1.116xx10^(-10) m`
=0.112 nm
Therefore , the de Broglie wavelength of the electron is 0.112 nm.