Animated Solution for Physics - Atoms and Nuclei: A particular hydrogen like ion emits radiation of frequency $2.92\times 10^{15}$ Hz when it makes transition from $n=3$ to $n=1$. The frequency in Hz of radiation emitted in transition from $n=2$ to $n=1$ will be

Imagine you are observing a tiny, invisible stage where an electron performs spectacular jumps. We are dealing with a hydrogen-like ion, which means it has a single electron orbiting a nucleus, much like a classic hydrogen atom but potentially with a stronger positive charge at the center.

Our electron is initially resting at the energy level $n=3$. Suddenly, it plunges down to the ground state, $n=1$. In doing so, it sheds its excess energy by emitting a photon. We are told that the frequency of this emitted light is $f_1=2.92\times 10^{15}\text{ Hz}$.

But the question poses a new scenario: What if the electron had jumped from $n=2$ to $n=1$ instead? What would be the frequency, $f_2$, of that photon?

To decode the frequencies of these quantum jumps, we rely on the legendary Rydberg formula. This formula beautifully connects the frequency of emitted radiation to the principal quantum numbers of the initial and final states.

The frequency $f$ is given by:

Here, $R$ is the Rydberg constant, $c$ is the speed of light, and $Z$ is the atomic number of our mysterious ion.

Let's apply this to our first transition ($n=3\to n=1$):

Now, let's set up the equation for the second transition ($n=2\to n=1$):

At this point, you might be worried. We don't know the atomic number $Z$! How can we possibly solve for $f_2$?

This is where the magic of ratios comes in. In physics, whenever you have two states of the same system described by equations with identical constants, dividing them is often the most elegant path forward.

Let's divide $f_1$ by $f_2$:

Notice how the Rydberg constant $R$, the speed of light $c$, and the unknown atomic number squared $Z^2$ completely cancel out! We are left with a pure, beautiful fraction:

We have successfully isolated the relationship between the two frequencies. Now, it's just a matter of simple algebra. We rearrange our ratio to solve for $f_2$:

Substitute the known value of $f_1$:

Rounding to two decimal places to match our options, we get $2.46\times 10^{15}\text{ Hz}$.

The profound takeaway here is that the ratio $\frac{32}{27}$ is a universal constant for these specific transitions across any hydrogen-like species in the universe. Whether it's Hydrogen, $H{e}^{+}$, or $L{i}^{2+}$, the relative frequencies of these jumps remain perfectly locked in harmony.