What is the
(a)Momentum
(b)speed ,and
(c )de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Explanation

Kinetic energy ,
`K=120 eV=120xx1.6xx10^(-19)J`
`=1.92xx10^(-17)J`
`=1.92xx10^(-17)J`
`h=6.63xx10^(-34)Js`
`m=9.1xx10^(-31)Kg`
(a)Momentum of electron,
`p=sqrt(2mK)`
`therefore p=sqrt(2xx9.1xx10^(-31)xx1.92xx10^(-17))`
`therefore p=sqrt(34.944xx10^(-48))`
`therefore p=5.91xx10^(-24) kgms^(-1)`
(b)Speed of electron,
`v=(p)/(m)=(5.91xx10^(-24))/(9.1xx10^(-31))`
`therefore v=0.649xx10^(7)`
`therefore v~~6.5xx10^(6)m//s`
(c )de-Broglie wavelength,
`lambda=(h)/(p)`
`therefore lambda=(6.63xx10^(-34))/(5.91xx10^(-24))`
`therefore lambda=1.121xx10^(-10)m`
`therefore lambda=0.1121 nm`