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

Answer

`4.9xx10^(-6)eV`

Explanation

Here, `lambda=5500Å =5500xx10^(-10)m`
`55xx10^(-8)m`
de-Broglie wavelength associated with electron is
`lambda=h/(sqrt(2m_(e)E)) or E=(h^(2))/(2m_(e)lambda^(2))=1/(2m_(e)) (h/lambda)^(2)`
`:. E=1/(2xx(9.1xx10^(-31)))[(6.6xx10^(-34))/(55xx10^(-8))]^(2)`
`=7.9xx10^(-25)J`
`=(7.9xx10^(-25))/(1.6xx10^(-19))=4.9xx10^(-6)eV`