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)

Options

A. `1`
B. `2`
C. `3`
D. `4`

Answer

B

Explanation

Here, `lambda=90nm, K=10.4 eV`
As `E_(n)=-13.6/(n^(2))eV`
Now `(hc)/lambda-(13.6 eV)/(n^(2))=10.4eV`
`(1242eV nm)/(90 nm)-(13.6eV)/(n^(2))=10.4`
or `4.14/3-13.6/(n^(2)) or 13.8-10.4=13.6/(n^(2))`
or `3.4=13.6/(n^(2)) or n^(2)=13.6/3.4 or n^(2)=4` to `n=2`