Animated Solution for Physics - Atoms and Nuclei: A gas of identical hydrogen-like atoms has some atoms in the lowest (ground) energy level $A$ and some atoms in a particular upper (excited) energy level $B$ and there are no atoms in any other energy level. The atoms of the gas make the transition to a higher energy level by absorbing monochromatic light of photon energy $2.7\text{ eV}$. Subsequently, the atoms emit radiation of only six different energy photons. Some of the emitted photons have an energy of $2.7\text{ eV}$, some have more energy and some less than $2.7\text{ eV}$. (a) Find the principal quantum number of the initially excited level $B$. (b) Find the ionization energy for the gas atoms. (c) Find the maximum and the minimum energies of the emitted photons.

The beauty of Bohr's atomic model lies in its discrete, quantized energy levels. This problem is a classic puzzle that tests our ability to deduce the initial and final states of an atom based on the spectrum of light it emits and absorbs. Let's break it down step by step.

The problem tells us that after absorbing a monochromatic light of $2.7\text{ eV}$, the gas atoms emit exactly six different energy photons. This is a crucial clue! The number of spectral lines emitted when an electron de-excites from a higher state $n_f$ to all possible lower states is given by the combination formula:

Equating this to 6, we get:

Solving this quadratic equation, we find that $n_f=4$. So, the atoms were excited to the $4^{\text{th}}$ energy level.

Now, we need to figure out where the atoms started before they absorbed the $2.7\text{ eV}$ photon. Let's call this initial state $n_i$. The energy difference between the $4^{\text{th}}$ state and this initial state is exactly $2.7\text{ eV}$.

The problem states that during de-excitation, some emitted photons have energy exactly equal to $2.7\text{ eV}$, some have more, and some have less. Let's test the possibilities: - If $n_i=1$: The absorption energy ($1\to 4$) is $2.7\text{ eV}$. During emission, the maximum energy jump is $4\to 1$, which would be $2.7\text{ eV}$. All other jumps ($4\to 3$, $3\to 1$, etc.) would have energy less than $2.7\text{ eV}$. This contradicts the condition that some photons have more energy. - If $n_i=3$: The absorption energy ($3\to 4$) is $2.7\text{ eV}$. During emission, the minimum energy jump from the $4^{\text{th}}$ state is $4\to 3$, which is $2.7\text{ eV}$. All other jumps ($4\to 1$, $3\to 1$, etc.) would have energy greater than $2.7\text{ eV}$. This contradicts the condition that some photons have less energy. - If $n_i=2$: The absorption energy ($2\to 4$) is $2.7\text{ eV}$. During emission, the jump $4\to 2$ gives exactly $2.7\text{ eV}$. The jump $4\to 3$ gives less energy, and the jump $4\to 1$ gives more energy. This perfectly matches all conditions!

Therefore, the initial excited state $B$ corresponds to the principal quantum number $n=2$.

The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen-like atom is given by:

where $E_0$ is the ionization energy of the atom. We know that the energy difference between the $4^{\text{th}}$ and $2^{\text{nd}}$ state is $2.7\text{ eV}$:

Substituting the energy formula:

So, the ionization energy of the gas atoms is $14.4\text{ eV}$.

The emitted photons correspond to all possible transitions from $n=4$ to lower states.

Maximum Energy: The maximum energy photon is emitted during the largest possible transition, which is from $n=4$ all the way down to the ground state $n=1$.

Minimum Energy: The minimum energy photon is emitted during the smallest possible transition, which is from $n=4$ to the adjacent state $n=3$.

This problem beautifully ties together the concepts of absorption, emission, and the mathematical structure of Bohr's energy levels!