Calculate the
momentum and
de-Broglie wavelength of the electrons accelerated through a potential difference of 56V.

Explanation

Mass of electorn `=m =9.1 xx 10^(-31)kg`
Potential difference= V=56V
de-Broglie wavelength of electron `=lamda`
`=(h)/(sqrt(2meV))`
`=(6.62 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(-31) xx 1.6 xx 10^(-19) xx 56))`
`=(6.62 xx 10^(-34))/(40.38 xx 10^(-25))`
`=0.1639 xx 10^(-9)m`
`lamda= 1.639 xx 10^(-10) m= 1.639 Å`
Momentum of electron `=P =(h)/(lamda)`
`=(6.62 xx 10^(-34))/(1.639 xx 10^(-10))`
`P= 4.029 xx 10^(-24)kg m//s`