Animated Solution for Physics - Atoms and Nuclei: In a hydrogen like atom electron make transition from an energy level with quantum number $n$ to another with quantum number $(n-1)$. If $n>>1$, the frequency of radiation emitted is proportional to

The journey of an electron transitioning between energy levels is one of the most fundamental and beautiful concepts in quantum mechanics. In this problem, we are asked to explore a specific transition in a hydrogen-like atom and determine how the frequency of the emitted radiation scales with the principal quantum number, $n$, especially when $n$ is extremely large.

This isn't just a dry algebraic exercise; it is a profound demonstration of how the bizarre rules of the quantum world smoothly blend into the familiar laws of classical physics—a concept famously known as Bohr's Correspondence Principle. Let's break down the physics and the math step-by-step.

Imagine an electron orbiting the nucleus of a hydrogen-like atom. It is currently residing in a highly excited state, denoted by the principal quantum number $n$. The problem states that this electron makes a transition to the immediate next lower energy level, which is $n-1$.

When an electron drops from a higher energy state to a lower one, it must shed the excess energy. It does this by emitting a photon—a tiny packet of light. According to the Bohr model, the energy of this emitted photon, $h u$, is exactly equal to the energy difference between the initial and final orbits.

To find the frequency $ u$ of this emitted photon, we rely on the Rydberg formula, which is derived directly from the energy level equations of the Bohr model. The frequency of radiation emitted during a transition from an initial state $n_i$ to a final state $n_f$ is given by:

For the sake of simplicity, let's bundle all the constants ($c$, $R$, and $Z^2$) into a single proportionality constant, $k$. Our equation then becomes:

In our specific scenario, the electron starts at $n_i=n$ and ends at $n_f=n-1$. Substituting these values into our master equation gives us the raw setup for our calculation:

Now, we need to roll up our sleeves and do some algebra. Our goal is to combine these two fractions into a single expression. We start by finding a common denominator, which is simply the product of the two individual denominators: $n^2(n-1{)}^2$.

Multiplying the numerators accordingly, we get:

Next, we expand the term $(n-1{)}^2$ in the numerator. Using the standard algebraic identity $(a-b{)}^2=a^2-2ab+b^2$, we have:

Substituting this back into our numerator:

Notice how beautifully the $n^2$ terms cancel each other out! Distributing the negative sign, we are left with:

This is the exact expression for the frequency of the emitted photon for any transition from $n$ to $n-1$.

Here is where the physics gets truly interesting. The problem provides a crucial condition: $n\gg 1$. This means $n$ is a very, very large number. We are dealing with an electron in a macroscopic orbit, far away from the nucleus.

When $n$ is massive, subtracting a tiny number like 1 makes virtually no difference. Think about it: if you have a million dollars and you lose one dollar, you essentially still have a million dollars.

Therefore, we can apply the following approximations: 1. In the numerator: $2n-1\approx 2n$ 2. In the denominator: $n-1\approx n$

Let's inject these approximations back into our simplified equation:

Simplifying the denominator, $n^2\cdot n^2$ becomes $n^4$. Our equation now looks like this:

We can cancel one factor of $n$ from both the numerator and the denominator, leading us to our final, elegant result:

Since $2k$ is just another constant, we can conclude that the frequency of the emitted radiation is directly proportional to the inverse cube of the principal quantum number:

This perfectly matches option (d).

The Grand Finale: Why is this result so special? In classical electrodynamics, an accelerating charge (like an electron orbiting a nucleus) should continuously emit radiation at a frequency equal to its frequency of revolution. If you calculate the classical frequency of revolution for an electron in the $n$-th Bohr orbit, you will find that it is also exactly proportional to $1/n^3$!

This is no coincidence. It is a direct manifestation of Bohr's Correspondence Principle, which states that for very large quantum numbers, the predictions of quantum mechanics must seamlessly converge with the predictions of classical physics. The math we just did proves that the quantum universe and the classical universe are, ultimately, one and the same.