An electron, in a hydrogen like atom , is in excited state. It has a total energy of -3.4 eV, find the de-Broglie wavelength of the electron.

Answer

(i) `3.4 eV` (ii) `6.63 Å`

Explanation

(i) Kinetic energy of electron in the orbits of hydrogen and hydrogen like atoms `=|"Total energy"|`
`:.` Kinetic energy `=3.4 eV`
(ii) The Be-Broglie wavelength is given by
`lambda=h/P=(h)/sqrt(2Km)`
K= kinetic energy of electron
substituting the values, we have
`lambda=((6.6xx10^(-34)J-s))/sqrt(2(3.4xx1.6xx10^(-19)J)(9.1xx10^(-31) kg))`
`lambda=6.63xx10^(-10) m` or `lambda=6.63 Å`