Animated Solution for Chemistry - Atomic Structure: The difference between the radii of 3rd and 4th orbits of ${\text{Li}}^{2+}$ is $\Delta R_1$. The difference between the radii of 3rd and 4th orbits of ${\text{He}}^{+}$ is $\Delta R_2$. Ratio $\Delta R_1:\Delta R_2$ is

Imagine you are looking at the atomic structure of an ion through a highly advanced microscope. According to Bohr's Model, electrons revolve around the nucleus in fixed circular paths called orbits.

The radius of any $n^{\text{th}}$ orbit is given by the elegant formula:

Here, $n$ is the principal quantum number (the orbit number), and $Z$ is the atomic number (the number of protons in the nucleus). Notice that as $n$ increases, the radius grows quadratically, meaning the orbits get progressively further apart!

The problem asks us to find the difference between the radii of the $4^{\text{th}}$ and $3^{\text{rd}}$ orbits. Let's denote this difference as $\Delta R$.

Using our radius formula, we can write:

Substituting the formula for $r_4$ and $r_3$, we get:

We can factor out the common terms to make our lives easier:

This simplified expression is our master key. It tells us that for a specific transition (like $n=3$ to $n=4$), the difference in radii is simply inversely proportional to the atomic number $Z$.

Now, let's apply our master key to the two ions given in the problem.

First, we have the Lithium ion (${\text{Li}}^{2+}$). For Lithium, the atomic number is $Z=3$. Substituting this into our equation gives us $\Delta R_1$:

Next, we have the Helium ion (${\text{He}}^{+}$). For Helium, the atomic number is $Z=2$. Substituting this gives us $\Delta R_2$:

A quick pro-tip: Notice how we didn't multiply $0.529$ by $7$? In competitive exams, always hold off on tedious calculations until the very end. Often, terms will cancel out!

We are finally ready to find the ratio $\Delta R_1:\Delta R_2$. Let's divide the two expressions we just found:

Just as we hoped, the bulky $0.529\times 7$ term appears in both the numerator and the denominator. They cancel out beautifully, leaving us with a simple fraction:

Simplifying this compound fraction, we get our final answer:

So, the ratio of the differences in radii is $2:3$. This elegant result highlights the power of algebraic simplification over brute-force calculation!