Animated Solution for Physics - Atoms and Nuclei: The first three spectral lines of H-atom in the Balmer series are given ${\lambda }_1,{\lambda }_2,{\lambda }_3$ considering the Bohr atomic model, the wavelengths of first and third spectral lines $\left(\frac{{\lambda }_1}{{\lambda }_3}\right)$ are related by a factor of approximately $x\times 10^{-1}$. The value of $x$ to the nearest integer, is ……… .

Imagine you are peering into the quantum world of a hydrogen atom. Electrons don't just float anywhere; they live in strict, quantized energy levels. When an excited electron decides to jump down to a lower energy level, it sheds its excess energy by firing off a photon of light.

The Balmer series is a very special collection of these jumps. It specifically describes any electron that falls from a higher energy level down to the second energy level ($n=2$). What makes the Balmer series so famous is that these specific jumps release photons with energies that correspond to visible light—the very colors we can see with our own eyes!

In this problem, we are asked to compare the wavelengths of the first and third spectral lines of this series. The 'first' line is the smallest jump, from $n=3$ to $n=2$. The 'third' line is a larger jump, from $n=5$ to $n=2$.

Rydberg Formula

To unlock the secrets of these wavelengths, we use the legendary Rydberg formula:

Here, $R$ is the Rydberg constant, $n_1$ is the lower energy level (which is always $2$ for the Balmer series), and $n_2$ is the higher energy level the electron is jumping from.

Let's set up the equation for the first spectral line, ${\lambda }_1$. The electron jumps from $n_2=3$ to $n_1=2$. Substituting these values into our formula:

Now, we just need to do a little fraction magic. Squaring the numbers gives us $\frac{1}{4}$ and $\frac{1}{9}$. Finding a common denominator of $36$, we get:

Next, we move to the third spectral line, ${\lambda }_3$. This time, the transition is from $n_2=5$ to $n_1=2$. Plugging these into the Rydberg formula:

Squaring the numbers gives us $\frac{1}{4}$ and $\frac{1}{25}$. The common denominator here is $100$. The numerator becomes $25-4$, which is $21$. So, we have:

The question asks for the ratio of ${\lambda }_1$ to ${\lambda }_3$. Notice a brilliant mathematical shortcut: $\frac{{\lambda }_1}{{\lambda }_3}$ is exactly the same as $\frac{1/{\lambda }_3}{1/{\lambda }_1}$. Let's divide the two expressions we just found:

Look at that! The Rydberg constant $R$ cancels out beautifully. We are left with a pure ratio of numbers:

Let's convert this fraction into a decimal. Dividing $756$ by $500$ gives exactly $1.512$.

The problem requires the answer in a very specific format: $x\times 10^{-1}$. So, we need to shift the decimal point one place to the right to balance the $10^{-1}$ multiplier:

Comparing this to our target format, we see that $x=15.12$. Rounding to the nearest integer, we arrive at our final answer: 15.

A quick takeaway: The ratio of any two wavelengths in a spectral series is completely independent of the Rydberg constant and the atomic number $Z$. Always look for these elegant cancellations in physics problems!