Calculate the (a) momentum and (b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56V. Given, `h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg, e=1.6xx10^(-19)C`.

Explanation

Potential difference , V=56 V
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)At equilibrium , the kinetic energy of each electron is equal to the accelerating potential , i.e., we can write the relation for velocity (v) of each electron as :
`1/2mv^2=eV`
`v^2=(2eV)/(m)`
`therefore v=sqrt((2xx1.6xx10^(-19)xx56)/(9.1xx10^(-31)))`
`=sqrt(19.69xx10^(12))=4.44xx10^6` m/s
The momentum of each accelerated electron is given as :
p=mv
`=9.1xx10^(-31) xx4.44 xx10^6`
`=4.04xx10^(-24) kg m s^(-1)`
Therefore, the momentum of each electron is `4.04xx10^(-24) kg m s^(-1)`
(b)De Broglie wavelength of an electron accelerating through a potential V , is given by the relation :
`lambda=12.27/sqrtVÅ`
`lambda=12.27/sqrt56 xx10^(-10) m `
=0.1639 nm
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.