What is the wavelength for the electron accelerated by ` 1.0 xx 10^(4) `volts?

Explanation

Step I. Calculating of the velocity of electron
Energy (kinetic energy ) of electron `=1.0 xx 10^(4)` volts.
`=1.0 xx 10^(4) xx 1.6 xx 10^(-19)J=1.6 xx 10^(-15)J`
`=1.6 xx 10^(-15)kg m^(2)s^(-2)`
or `" " 1//2 m upsilon^(2)=1.6 xx 10^(-15)kg m^(2)s^(-2)`
or `" " upsilon =((2xx1.6 xx10^(-15)kg m^(2)s^(-2))/(9.1 xx 10^(-31)kg))^(1//2)=5.93 xx 10^(7) ms^(-1)`
Step II. calculation of the wavelength of electron
According to de Broglie equation,
`lambda =(h)/(m upsilon) , lambda = ((6.626 xx 10^(-34) kg m ^(2) s^(-1)))/((9.1 xx 10^(-31 ) kg) xx (5.93 xx 10^(7) ms^(-1)))=1.22 xx 10^(-11)m`.