Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom ?
(A) The ra

Q: Why does the longest wavelength correspond to the transition from the very next energy level?

According to Planck's equation, $E = \frac{hc}{\lambda}$, energy and wavelength are inversely proportional. The smallest energy difference between levels occurs when an electron jumps from the immediately adjacent higher level (e.g., $n=3$ to $n=2$). Smallest energy means the longest possible wavelength.

Q: What does $n = \infty$ physically mean in these transitions?

An energy level of $n = \infty$ means the electron is completely free from the nucleus's electrostatic pull; the atom is ionized. A transition from $n = \infty$ to a lower level means a free electron is captured by the nucleus, releasing the maximum possible energy for that series.

Q: Why is there no overlap between the Lyman and Balmer series, but higher series do overlap?

The energy gaps between lower quantum numbers (like $n=1$ and $n=2$) are massive compared to the gaps between higher quantum numbers. The entire energy range of the Balmer series fits within a small fraction of the Lyman series' energy gap, preventing any overlap. As $n$ increases, the levels get closer together, causing higher series like Brackett and Pfund to overlap.

Q: In Option C, why is the given formula wrong?

The given formula incorrectly uses a plus sign and multiplication. If you derive it properly using $\frac{1}{\lambda_0} = R$, the correct relation is $\lambda = \frac{\lambda_0}{1 - \frac{1}{m^2}}$. The option's formula would imply wavelengths longer than what is physically possible for the Lyman series.

Q: Do I need to memorize the exact wavelength values to solve this?

Not at all! As shown in the solution, you only need to work with the expressions in terms of the Rydberg constant $R$. By comparing the coefficients of $1/R$, you can easily determine ratios and check for overlaps without calculating a single numerical wavelength.

Question

Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom ?

Options:

(A) The ratio of the longest wavelength to the shortest wavelength in Balmer series is $\frac{9}{5}$

(B) There is an overlap between the wavelength ranges of Balmer and Paschen series.

(C) The wavelengths of Lyman series are given by $\left(1 + \frac{1}{m^2}\right)\lambda_0$, where $\lambda_0$ is the shortest wavelength of Lyman series and $m$ is an integer

(D) The wavelength ranges of Lyman and Balmer series do not overlap

Sigma Percentile · Accepted Answer

### The Quantum Amphitheater

Imagine you are standing in a grand quantum amphitheater, watching an electron perform spectacular leaps between the energy levels of a hydrogen atom. Every time it jumps down to a lower level, it releases a burst of energy in the form of a photon. The color, or wavelength, of this photon depends entirely on the height of the jump.

To unlock the secrets of these spectral lines, we rely on a single, elegant equation known as the Rydberg formula.

### The Master Key: Rydberg's Formula

The Rydberg formula is our map to the hydrogen spectrum. It relates the wavelength $\lambda$ of the emitted photon to the initial and final energy levels of the electron:

$$ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) $$

Here, $R$ is the Rydberg constant, $n_1$ is the lower energy level (the destination), and $n_2$ is the higher energy level (the starting point). Remember a crucial physical principle: **energy is inversely proportional to wavelength**. A massive energy jump produces a tiny, short wavelength, while a small, lazy hop produces a long, stretched-out wavelength.

### Option A: The Balmer Series Boundaries

Let's dive into the Balmer series, where the electron always falls back to the second energy level ($n_1 = 2$).

To find the **longest wavelength** ($\lambda_{\max}$), we need the smallest possible energy jump. This occurs when the electron hops down from the very next level, $n_2 = 3$. Plugging this into our formula:

$$ \frac{1}{\lambda_{\max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36} $$

Flipping this over, we get $\lambda_{\max} = \frac{36}{5R}$.

Conversely, the **shortest wavelength** ($\lambda_{\min}$) corresponds to the most extreme jump possible—from the edge of the universe, $n_2 = \infty$, down to $n_1 = 2$.

$$ \frac{1}{\lambda_{\min}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} ight) = \frac{R}{4} $$

This gives us $\lambda_{\min} = \frac{4}{R}$. Now, let's find their ratio:

$$ 	ext{Ratio} = \frac{\lambda_{\max}}{\lambda_{\min}} = \frac{\frac{36}{5R}}{\frac{4}{R}} = \frac{9}{5} $$

This perfectly matches the statement in Option A. It is absolutely correct!

### Option B: Do Balmer and Paschen Overlap?

What does it mean for two spectral series to overlap? It means the longest, lowest-energy wavelength of the lower series stretches far enough to cross into the shortest, highest-energy wavelength of the upper series.

Let's check the Paschen series, where electrons fall to $n_1 = 3$. Its shortest wavelength occurs for a jump from $n_2 = \infty$:

$$ \frac{1}{\lambda_{\min, P}} = R \left( \frac{1}{3^2} - 0 ight) = \frac{R}{9} \implies \lambda_{\min, P} = \frac{9}{R} $$

We already know the longest wavelength of the Balmer series is $\lambda_{\max, B} = \frac{36}{5R} = \frac{7.2}{R}$.

Comparing the two, $\frac{7.2}{R}$ is strictly less than $\frac{9}{R}$. There is a clear, empty gap between the two series. They do not overlap, making Option B incorrect.

### Option C: Decoding the Lyman Formula

Now, let's turn our attention to the Lyman series, the most energetic series where electrons plummet all the way down to the ground state, $n_1 = 1$. The general formula is:

$$ \frac{1}{\lambda} = R \left( 1 - \frac{1}{m^2} ight) $$

The shortest wavelength, $\lambda_0$, happens when $m = \infty$. Substituting this gives $\frac{1}{\lambda_0} = R$.

If we substitute $R$ back into the general equation, we get:

$$ \frac{1}{\lambda} = \frac{1}{\lambda_0} \left( 1 - \frac{1}{m^2} ight) $$

Rearranging for $\lambda$, we find:

$$ \lambda = \frac{\lambda_0}{1 - \frac{1}{m^2}} $$

The formula provided in Option C uses a plus sign and multiplication, which is mathematically flawed. Option C is incorrect.

### Option D: The Lyman-Balmer Gap

Finally, we check if the Lyman and Balmer series overlap. We need the longest wavelength of the Lyman series, which is a jump from $n_2 = 2$ to $n_1 = 1$:

$$ \frac{1}{\lambda_{\max, L}} = R \left( 1 - \frac{1}{2^2} ight) = \frac{3R}{4} \implies \lambda_{\max, L} = \frac{4}{3R} \approx \frac{1.33}{R} $$

We compare this to the shortest wavelength of the Balmer series, which we found earlier to be $\lambda_{\min, B} = \frac{4}{R}$.

Since $\frac{1.33}{R}$ is vastly smaller than $\frac{4}{R}$, the Lyman series ends long before the Balmer series even begins. They do not overlap. Option D is correct!

### The Final Verdict

After a rigorous mathematical journey through the energy levels of hydrogen, we have systematically proven that only statements **(A)** and **(D)** hold true. This problem beautifully illustrates how algebraic manipulation of the Rydberg formula can reveal the hidden structure of atomic spectra.

Animated Solution for Physics - Atoms and Nuclei: Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom ?

Select Answer:

Visualized Solution

Hydrogen Spectrum

Rydberg Formula

Balmer Series Setup

Longest Wavelength $λ_{m a x}$

Shortest Wavelength $λ_{m i n}$

Ratio for Option A

Paschen Series Setup

Checking Overlap: Balmer vs Paschen

Lyman Series Setup

Lyman Wavelength Formula

Longest Wavelength $λ_{m a x}$ for Lyman

Checking Overlap: Lyman vs Balmer

Final Conclusion

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Quantum Amphitheater

The Master Key

Option A

Option B

Option C

Option D

The Final Verdict

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