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

`(hc)/(lambda)-{13.6eV.(1)/(2)}=10.4`
`rArr(124eV)/(90)-(13.6)/(n^(2))=10.4`
`rArr(41.4)/(3)-(13.6)/(n^(2))=10.4`
`rArr13.8-10.4=(13.6)/(n^(2))`
`rArr3.4=(13.6)/(n^(2))`
`rArrn^(2)=4`
`rArr n=2`