A sample of hydrogen atoms containing all the atoms in a particular excited state, absorb radiations — Atomic Structure Chemistry Question
Question
A sample of hydrogen atoms containing all the atoms in a particular excited state, absorb radiations of a particular wavelength by which the atoms get excited to another excited state. When the atoms finally de-excite to the ground state, they emit the radiations of 10 different wavelengths. Out of these 10 radiations, 7 have wavelengths shorter than the absorbed radiation and 2 have wavelength longer than the absorbed radiation. The orbit number for the initial excited state of atoms is
💡 Solution & Explanation
Total emitted lines is 10, meaning $\frac{n(n-1)}{2} = 10 \implies n = 5$. The final excited state is $n=5$. The atoms absorbed radiation to jump from $n_i$ to $n=5$. Since 7 emitted lines have shorter wavelengths (higher energy) and 2 have longer wavelengths (lower energy) than the absorbed photon, the absorbed photon is the 8th highest energy transition among the 10. Listing the transitions by energy: $5\to 4$ and $4\to 3$ are the lowest energies. The next lowest is $5\to 3$. If the absorbed photon corresponds to $5\to 3$ (energy 0.97 eV), it perfectly leaves 2 transitions with lower energy and 7 with higher energy. Thus, the transition is from $n=3$ to $n=5$, meaning $n_i = 3$. Therefore, correct answer is 3.