Animated Solution for Physics - Atoms and Nuclei: Comprehension Passage

Imagine a microscopic game of billiards, but instead of solid balls, we are dealing with atoms and ions exchanging pure packets of energy. In this fascinating scenario, we have a mixture of Hydrogen (H) atoms and singly ionized Helium (He${}^{+}$) ions. Both are initially resting in their first excited states.

What does "first excited state" mean? The lowest possible energy level, the ground state, corresponds to the principal quantum number $n=1$. The very first step up the energy ladder is $n=2$. Therefore, both our H atoms and He${}^{+}$ ions start their journey at $n=2$.

The Energy Exchange

To understand the energy transfer, we rely on the master equation of the Bohr model for hydrogen-like atoms:

For the Hydrogen atom ($Z=1$), the energy in the first excited state ($n=2$) is:

When the Hydrogen atom de-excites, it drops back down to its ground state ($n=1$), where its energy is $-13.6\text{ eV}$. The energy it releases during this drop is the difference between these two states:

This exact packet of $10.2\text{ eV}$ is transferred to the waiting He${}^{+}$ ion via a collision.

Now, let's look at the He${}^{+}$ ion ($Z=2$). It is also initially sitting at $n=2$. Let's calculate its initial energy:

Upon absorbing the $10.2\text{ eV}$ packet from the Hydrogen atom, its new total energy becomes:

We must now determine which quantum state $n$ corresponds to an energy of $-3.4\text{ eV}$ for a Helium ion. Using our master formula again:

Solving for $n^2$, we get $n^2=16$, which means $n=4$. The Helium ion has been excited to the $n=4$ state! This beautifully answers our first question.

Emitting Visible Light

What goes up must come down. The He${}^{+}$ ion, now at $n=4$, will eventually de-excite. We are specifically looking for a transition that emits a photon in the visible region of the electromagnetic spectrum. Visible light roughly corresponds to photon energies between $1.8\text{ eV}$ (red) and $3.1\text{ eV}$ (violet).

Let's test the possible downward jumps from $n=4$: - Jumping to $n=1$: $\Delta E=-3.4-(-54.4)=51.0\text{ eV}$ (Deep Ultraviolet) - Jumping to $n=2$: $\Delta E=-3.4-(-13.6)=10.2\text{ eV}$ (Ultraviolet) - Jumping to $n=3$: $\Delta E=-3.4-\left(-13.6\frac{4}{9}\right)=-3.4-(-6.04)=2.64\text{ eV}$

Bingo! The energy $2.64\text{ eV}$ falls perfectly within the visible spectrum. To find the exact wavelength of this light, we use the Planck-Einstein relation:

Using $hc\approx 1240\text{ eV nm}$:

Converting this to meters, we get $4.697\times 10^{-7}\text{ m}$, which is approximately $4.8\times 10^{-7}\text{ m}$, matching our options perfectly.

Finally, we are asked to compare the kinetic energies of the electrons in the $n=2$ state for both the H atom and the He${}^{+}$ ion. In the Bohr model, the kinetic energy $K$ is simply the positive magnitude of the total energy:

Since both electrons are in the $n=2$ orbit, their kinetic energies are directly proportional to the square of their atomic numbers ($Z^2$).

For Hydrogen ($Z=1$): $K_H\propto 1^2=1$ For Helium ion ($Z=2$): $K_{{\text{He}}^{+}}\propto 2^2=4$

Therefore, the ratio of their kinetic energies is simply $1/4$.

This problem is a masterclass in applying the Bohr model, seamlessly weaving together energy conservation, atomic transitions, and proportional reasoning.