Animated Solution for Chemistry - Atomic Structure: The frequency of light emitted for the transition $n=4$ to $n=2$ of ${\text{He}}^{+}$ is equal to the transition in H atom corresponding to which of the following?

Imagine two different musical instruments playing the exact same note. In the quantum world, atoms can do something similar. They can emit photons of the exact same frequency, even if they are entirely different elements.

In this problem, we are looking for a "quantum resonance" between a Helium ion (${\text{He}}^{+}$) and a Hydrogen atom ($\text{H}$). We need to find a transition in Hydrogen that perfectly matches the energy gap of the $n=4\to n=2$ transition in Helium.

To find the frequency of the emitted light, we rely on Rydberg's Formula. This powerful equation connects the frequency of the emitted photon to the atomic number and the quantum energy levels:

Here, $Z$ is the atomic number, $n_2$ is the higher energy level, and $n_1$ is the lower energy level.

Let's first calculate the frequency emitted by the ${\text{He}}^{+}$ ion. For Helium, the atomic number is $Z=2$. The electron is falling from the fourth orbit ($n_2=4$) to the second orbit ($n_1=2$).

Substituting these values into our formula:

Let's simplify the math. The $2^2$ becomes $4$. Inside the bracket, we have $\frac{1}{4}-\frac{1}{16}$, which simplifies to $\frac{3}{16}$.

This is the exact frequency of the "note" that the Helium ion is playing.

Now, let's look at the Hydrogen atom. For Hydrogen, the atomic number is $Z=1$. We don't know the energy levels yet, so let's keep them as $n_2$ and $n_1$.

According to the problem, this frequency must be exactly equal to the frequency emitted by the Helium ion. Let's equate them!

The $Rc$ terms cancel out beautifully on both sides, leaving us with a simple algebraic puzzle:

We need to find two integers, $n_1$ and $n_2$, that satisfy this equation. By simple observation, we can see that if we set $n_1=1$ and $n_2=2$, the equation holds true:

Thus, the corresponding transition in the Hydrogen atom is from $n=2$ to $n=1$.

While the calculation is straightforward, there is a much faster way to solve this using pure physical intuition.

We know that the energy of an electron in a Bohr orbit is proportional to $\frac{Z^2}{n^2}$.

For two different atoms to have the exact same energy levels (and thus the same energy gaps), the ratio $\frac{Z}{n}$ must remain constant!

Since $Z_{\text{He}}=2$ and $Z_{\text{H}}=1$, we get:

This means the principal quantum number for Hydrogen will always be exactly half of that for Helium. Half of $4$ is $2$, and half of $2$ is $1$. Therefore, the $4\to 2$ transition in Helium perfectly mirrors the $2\to 1$ transition in Hydrogen!