The ionisation energy of hydrogen atom is given to be 13-6 eV. A photon falls a hydrogen atom which is initially in the ground state and excites it to the n = 4 state.Calculate the wavelength of the photon.

Explanation

Energy of electron in the first orbit of hydrogen atom = 13-6 eW =`E_(1)` Energy of electron in the fourth orbit of `E_(4)=(13.6)/(4)^(2)`
`therefore` Energy required to excite an electron from the first orbit of hydrogen atom to fourth orbit =E
We know E=`(hc )/(lambda)`
where `lambda` = wavelength of incident photon
`lambda=(hc)/(E)`
`=973.5 xx10^(10) m`
`lambda=973.5 A`