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.

Explanation

a. energy of electron in hydrogen-like atom,
`E_(n) = - (Z^(2) Rhc)/(n^(2)) = - 3.4 e V`
Kinetic energy of electron in hydrogen-like atom is equal to negative of total energy
i.e. , ` E_(k) = - E_(n) = - (-3.4 e V) = + 3.4 e V`
(ii) The de Broglie wavelength of electron .
`lambda = (h)/(p) = (h)/sqrt(2 mE_(k))`
`= (6.63 xx 10^(-34))/sqrt{{2 xx 9.1 xx 10^(-31) xx 3.4 xx 1.6 xx 10^(-19)}}
`= 6.66 Å`