Question

A hydrogen-like atom in ground state absorbs n photons having the same energy and its emits exactly n photons when electrons transition takes placed. Then, the energy of the absorbed photon may be :

• A. $91.8eV$
• B. $40.8eV$
• C. $48.4eV$
• D. $54.4eV$

Answer 1

Answer: (A), (B)

The correct options are
A $91.8eV$
B $40.8eV$

Number of dark lines(in absorption) i.e excitation = number of bright lines(in emission),i.e de-excitation
It is possible only when the e− is excited to $n=2$ from ground state
Clearly, ΔE = 91.8 eV and 40.8 eV are possible
The energy of electron in the first and second orbit is $-13.6eV$ and $-3.4eV$.
The energy difference is $13.6-3.4=10.2$ eV.

(A) $\frac{40.8}{10.2}=4$ an integer

(B) $\frac{91.8}{10.2}=9$ an integer