If the kinetic energy of an electron is `4.55xx10^(-25)J`, find its wavelength (Planck's constant, `h=6.6xx10^(-34)kgm^(2)s^(-1),m=9.1xx10^(-31)kg)`.

Explanation

Kinetic elergy (KE) of electron, `KE=(1)/(2)mv^(2)=4.55xx106(-25)J`, Mass of electron, `m=9.1xx10^(-31)kg`
`KE=(1)/(2)mv^(2), mv^(2)=2KE (or) m^(2)v^(2)=2mKE`
`mv=(h)/(sqrt(2mKE)) and lambda=(h)/(mv)=(h)/(sqrt(2mKE))`
Wave length `lambda=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx4.55xx106(-25)))=7.25xx10^(-7)m=725nm`