Consider a hydrogen atom with its electron in the `n^(th)` orbital An electomagnetic radiation of wavelength `90 nm` is used to ionize the atom . If the kinetic energy of the ejected electron is `10.4 eV` , then the value of `n` is ( hc = 1242 eVnm)

Answer

[2]

Explanation

Energy required to liberate electron from `n^(the)` orbit of hydrogen atom is given by
`E_n=(13.6)/n^2eV`
Energy of the incident photon can be written as follows :
`E_p=(hc)/lamda=1242/90 = (eVnm)/(nm)=13.8eV`
Kinetic energy of the ejected is given as follows :
K.E. = 10.4 eV
When electron is ejected , then remaining nucleus recoils . But nucleus being very heavy in comparison of electron, we can neglect recoil energy of the nucleus. Hence by conservation of energy , we can write the following expression.
`13.8eV = (13.6)/n^2eV+10.4eV`
`impliesn^2=(13.6)/(3.4)=4`
`implies n=2`