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

D

Explanation

(d) `KE_1 = E_("photon") - BE_(n = 1)`
`KE_2 = E_("photon") - BE_(n = n)`
`KE_2 - KE_1 = BE_(n =1) - BE_(n =n)`
=`13.6 Z^2[(1)/(1^2) - (1)/(n^2)] = 12.75`(given)
`:. n_2 = 16`
or `n = 4`
`BE` : Binding energy.