In a hydrogen like atom, electron is in `2^(nd)` excited state, the binding energy of fourth state of this atom is 13.6 eV, then

Options

A. A 25 eV photon can set free the electron from the second excited state of this sample.
B. 3 different types of photon will be observed if electrons make transition up to ground state from the second excited state
C. If 23 eV photon is used then K.E. of the ejected electron is 1eV.
D. `2^(nd)` line of balmer series of this sample has same energy value as `1^(st)` excitation energy of H-atoms

Answer

A, B

Explanation

B.E. of `4^(th)` state
`=13.6 (z^2)/(n^2) implies 13.6 (z^2)/(4^2) = 13.6 implies z= 4`
Sample is `Be^(3+) therefore` energy of electron in `3^(rd)` state
Therefore 25 eV photon will cause ionization