The de Broglie wavelength of an electron with kinetic energy 120 e V is `("Given h"=6.63xx10^(-34)Js,m_(e)=9xx10^(-31)kg,1e V=1.6xx10^(-19)J)`

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 .

if the de broglie wavelength of an electron is 0.3 nanometre, what is its kinetic energy ? [h=6.6xx10^(-34) Js, m=9xx10^(-31)kg, 1eV = 1.6xx10^(-19) J]

If de Broglie wavelength of an electron is 0.5467 Å, find the kinetic energy of electron in eV. Given h=6.6xx10^(-34) Js , e= 1.6xx10^(-19) C, m_e=9.11xx10^(-31) kg.

Calculate the de-Broglie wavelength of an electron of kinetic energy 100 eV. Given m_(e)=9.1xx10^(-31)kg, h=6.62xx10^(-34)Js .

Photon having energy equivalent to the binding energy of 4th state of He^(+) ion is used to eject an electron from the metal with KE/2eV . If electron is further accelerated through a potential difference of 4V then the minimum value of de Broglie wavelength associated with the electron is : (h = 6.6 xx 10^(-34) J-s , m_(e) = 9.1 xx 10^(-31) kg . 1 eV = 1.6 xx 10^(-19) J )

The de Broglie wavelength of an electron in a metal at 27^(@)C is ("Given"m_(e)=9.1xx10^(-31)kg,k_(B)=1.38xx10^(-23)JK^(-1))

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 .

Find de-Broglie wavelength of electron with KE =9.6xx10^(-19)J .

Calculate the energy of an electron having de-Broglie wavelength 5500Å. Given, h=6.62xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg .

De - Broglie wavelength of an electron accelerated by a voltage of 50 V is close to (|e|=1.6xx10^(-19)C,m_(e)=9.1xx10^(-31))