Animated Solution for Physics - Atoms and Nuclei: In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be

Decoding Wavelengths in Hydrogen-like Atoms

Imagine you are an electron orbiting a nucleus. You aren't allowed to just float anywhere; you must reside in specific, quantized energy levels known as shells. When you drop from a higher energy shell to a lower one, you shed your excess energy by emitting a photon of light. This beautiful phenomenon is the heart of atomic spectra, and in this problem, we are going to decode the exact wavelengths of these emitted photons.

Shells and Quantum Numbers

Before we dive into the math, we need to translate the alphabetical names of the shells into the language of physics: principal quantum numbers ($n$).

The naming convention starts from the innermost shell and moves outward alphabetically: - K-shell corresponds to $n=1$ - L-shell corresponds to $n=2$ - M-shell corresponds to $n=3$ - N-shell corresponds to $n=4$

Our master key for this problem is the Rydberg Formula, which elegantly connects the wavelength of the emitted photon ($\lambda$) to the initial and final energy levels:

Here, $R$ is the Rydberg constant, $Z$ is the atomic number, $n_1$ is the lower energy level (where the electron lands), and $n_2$ is the higher energy level (where the electron starts).

M to L Shell

In our first scenario, the electron takes a leap from the M-shell ($n_2=3$) down to the L-shell ($n_1=2$). The problem tells us that this specific jump emits a photon with a wavelength of $\lambda$.

Let's plug these values into our master equation:

Squaring the numbers, we get:

Taking the common denominator (which is 36), we simplify the expression:

N to L Shell

Now, let's look at the second scenario. The electron is feeling a bit more energetic and jumps from an even higher ledge: the N-shell ($n_2=4$) down to the same L-shell ($n_1=2$). Let's call the wavelength of this newly emitted photon ${\lambda }^{\prime }$.

Applying the Rydberg formula again:

Finding the common denominator (which is 16 this time), we get:

Comparing the Wavelengths

We now have two equations, but we don't know the exact values of $R$ or $Z$. However, we don't need to! Since we are looking for a relationship between ${\lambda }^{\prime }$ and $\lambda$, we can simply divide Equation 1 by Equation 2. Notice the elegance of the cancellation; the $R{Z}^2$ terms vanish completely!

This simplifies to:

Let's do some quick cross-cancellation. Both 16 and 36 are divisible by 4:

Finally, isolating ${\lambda }^{\prime }$, we arrive at our answer:

Does this mathematical result make physical sense? Absolutely.

The energy gap between the 4th shell and the 2nd shell is larger than the gap between the 3rd shell and the 2nd shell ($E_{4\to 2}>E_{3\to 2}$). Because the energy of a photon is inversely proportional to its wavelength ($E=\frac{hc}{\lambda }$), a larger energy drop must result in a shorter wavelength.

Our result, ${\lambda }^{\prime }=\frac{20}{27}\lambda$, shows that ${\lambda }^{\prime }$ is indeed smaller than $\lambda$ (since $\frac{20}{27}<1$). The math perfectly aligns with the physical reality of the quantum world!