Calculate the wavelength of the wave associated with an electron beam, if the beam is accelerated by a potential difference of 5000 volt.

Explanation

Kinetic energy of the electron=5000 eV
`=5000xx1.602xx10^(-19)J`
Velocity of an electron due to the applied potential difference is `vms^(-1)`. Hence, kinetic energy`=(1)/(2)mv^(2)`.
`therefore (1)/(2)mv^(2)=5000xx1.602xx10^(-19)J`
`=5000xx1.602xx10^(-19)kg*m^(2)*s^(-1)`
`therefore v=sqrt((2xx5000xx1.602xx10^(-19))/(9.11xx10^(-31)))=4.193xx10^(7)m*s^(-1)" "[because"mass of electron"=9.11xx10^(-31)kg]`
From de Broglie equation, we get, `lamda=(h)/(mv)`
`=(6.626xx10^(-34))/(9.11xx10^(-31)xx4.193xx10^(7))`
`=0.1736xx10^(-10)m=0.1736` Å