Animated Solution for Physics - Atoms and Nuclei: In a hydrogen atom, electron makes a transition from $(n+1)$th level to the $n$th level. If $n>>1$, the frequency of radiation emitted is proportional to

Imagine you are observing a hydrogen atom. An electron is currently residing in a highly excited state, specifically the $(n+1)$th energy level. Suddenly, it makes a quantum leap down to the adjacent lower level, the $n$th level.

Whenever an electron drops to a lower energy state, it must shed the excess energy. It does this by emitting a photon. Our goal is to find out how the frequency of this emitted photon relates to the principal quantum number $n$, especially when $n$ is a very large number.

To find the frequency, we rely on Bohr's frequency condition. The energy of the emitted photon, given by $E=hf$ (where $h$ is Planck's constant and $f$ is the frequency), is exactly equal to the difference in energy between the initial and final states.

We know from Bohr's model that the energy of an electron in the $n$th orbit of a hydrogen atom is:

Let's substitute this into our frequency equation. We must be careful with our negative signs!

Now, let's clean up this expression. We can factor out the common term, $\frac{13.6}{h}$, and rearrange the fractions to make them positive.

To subtract these fractions, we need a common denominator, which will be $n^2(n+1{)}^2$.

Let's expand the numerator using the identity $(a+b{)}^2=a^2+2ab+b^2$. The $n^2$ terms will beautifully cancel out.

Here is where the physics intuition kicks in. The problem states a crucial condition: $n\gg 1$. This means $n$ is a very large number.

When $n$ is massive, adding $1$ to it barely changes its value. Think about it: if you have a million dollars, finding one more dollar doesn't change your financial status significantly. Therefore, we can make the following approximations:

Let's substitute these approximations back into our simplified frequency equation.

We are almost there! We can cancel one $n$ from the numerator with one $n$ from the $n^4$ in the denominator.

Since $13.6$, $2$, and $h$ are all constants, we can conclude that the frequency $f$ is directly proportional to $\frac{1}{n^3}$.

This elegant result is a perfect illustration of Bohr's Correspondence Principle, which states that for very large quantum numbers, quantum physics yields the same results as classical physics!