An electron in H-atom in its ground state absorbs ` 1.5` times as much energy as the minimum required for its escape ( i. e., 13 . 6 eV) from the atom . Calculate the wavelength of emitted electron.

Explanation

For the electrons, `E_(1)=13.6e" and "E_(2)=13.6xx1.50=20.4eV`
`DeltaE=(E_(2)-E_(1))=20.4-13.6=6.8eV`
`6.8xx1.6xx10^(-19)J=10.88xx10^(-19)kg m^(2)s^(-2)`
The energy difference `(DeltaE)` represents the energy gained by the electron as kinetic energy
Now, `lambda=(h)/(sqrt(2m(KE)))=((6.626xx10^(-34)kgm^(2)s^(-1)))/(sqrt((2xx9.108xx10^(-31)kgxx10.88xx10^(-19)kgm^(2)s^(-2)))=4.71xx10^(-10)m`