Photons of equal energy were incident on two different gas samples. One sample containing H-atoms in the ground state and the other sample containing H-atoms in some excited state with a principle quantum number 'n'. The photonic beams totally ionise the H-atoms. If the difference in the kinetic energy of the ejected electrons in the two different cases is `12.75 eV`. Then find the principal quantum number 'n' of the excited state.

Options

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

Answer

4

Explanation

`KE_(1)=E_("photon")-BE_(n+1)`
`KE_(2)=E_("photon")-BE_(n=n)`
`KE_(2)-KE_(1)=BE_(n=1)-BE_(n+n)=13.6Z^(2)[(1)/(1^(2))-(1)/(n^(2))]=12.75("given")`
`:. " "n^(2)=16" or "n=4`
`BE:` Binding energy.