Animated Solution for Physics - Atoms and Nuclei: If ${\lambda }_1$ and ${\lambda }_2$ are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of ${\lambda }_1:{\lambda }_2$ is

Welcome to the fascinating world of atomic physics! Today, we are going to dive deep into the hydrogen atom and explore the beautiful mathematics behind its emission spectra. Imagine the hydrogen atom as a microscopic building with a very specific set of floors, which we call energy levels. Electrons can jump between these floors, and when they jump down, they release energy in the form of light. Our mission is to find the ratio of the wavelengths of two very specific jumps.

Before we touch any equations, we must visualize the physical reality of the problem. The question mentions two distinct series of spectral lines: the Lyman series and the Paschen series.

Think of the Lyman series as the ground floor of our atomic building. Any electron that jumps down to the $n=1$ level belongs to the Lyman series. The "members" of the series simply tell us how many floors up the electron started. The first member is a jump from $n=2$ to $n=1$. The second member is from $n=3$ to $n=1$. Therefore, the third member of the Lyman series is a jump from $n=4$ to $n=1$. This jump emits a photon with wavelength ${\lambda }_1$.

Now, let's look at the Paschen series. This series represents all jumps down to the third floor, $n=3$. The first member of the Paschen series is the shortest possible jump down to this level, which is from $n=4$ to $n=3$. This jump emits a photon with wavelength ${\lambda }_2$.

Rydberg's Formula

To calculate the wavelengths of these emitted photons, we rely on a powerful tool known as the Rydberg formula:

Here, $R$ is the Rydberg constant, $n_1$ is the lower energy level, and $n_2$ is the higher energy level. Notice that the formula gives us the reciprocal of the wavelength ($1/\lambda$), which is directly proportional to the energy of the transition.

Let's apply the Rydberg formula to our two specific transitions.

For the third member of the Lyman series (${\lambda }_1$), the electron falls from $n_2=4$ to $n_1=1$. Substituting these values into our master equation, we get:

Similarly, for the first member of the Paschen series (${\lambda }_2$), the electron falls from $n_2=4$ to $n_1=3$. Plugging these into the formula yields:

We are asked to find the ratio ${\lambda }_1:{\lambda }_2$. A common trap here is to try and calculate the exact numerical value of each wavelength. That is a path filled with messy decimals and wasted time!

Instead, we can use a clever algebraic trick. Notice that if we divide the expression for $1/{\lambda }_2$ by the expression for $1/{\lambda }_1$, we get exactly what we want:

Let's set up this division. The beauty of this method is that the Rydberg constant $R$ appears in both the numerator and the denominator, allowing us to cancel it out completely!

Now, we just need to carefully simplify the fractions. Let's evaluate the squares first:

To subtract the fractions in the numerator, we find a common denominator, which is $9\times 16=144$.

The denominator is much simpler:

Putting it all together, we have a fraction divided by a fraction:

To divide by a fraction, we multiply by its reciprocal:

We can simplify this by noticing that $144$ is exactly $9\times 16$. So, the $16$ in the numerator cancels out with part of the $144$ in the denominator, leaving a $9$ behind:

And there we have it! The ratio of the wavelengths is exactly $7:135$.

This problem beautifully demonstrates the power of setting up ratios to eliminate constants and simplify calculations. Whenever you face a problem involving spectral series, always start by sketching the energy levels. It grounds your mathematics in physical reality and ensures you never pick the wrong quantum numbers. Keep practicing, and these atomic transitions will become second nature to you!