Given below are two statements : Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies. Statement II : The frequency of second line of Balmer series obtained from text{He}^+ is equal to that of first line of Lyman series obtained from hydrogen atom. In the light of the above statements, choose the correct answer from the options given below :

Hydrogen Spectrum Discrete Frequencies and Balmer vs Lyman Series Frequency Comparison

Q. Given below are two statements :

Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies.

Statement II : The frequency of second line of Balmer series obtained from $\text{He}^+$ is equal to that of first line of Lyman series obtained from hydrogen atom.

In the light of the above statements, choose the correct answer from the options given below :

(A) Both Statement I and Statement II are true

(B) Statement I is true but Statement II is false

(C) Statement I is false but Statement II is true

(D) Both Statement I and Statement II are false

Correct Answer: A

Both Statement I and Statement II are true

Explanation

Statement I describes the fundamental process of atomic emission. When high-voltage electric discharge passes through $H_2$ gas, the molecules absorb energy and dissociate into atoms. These atoms are excited to higher energy levels. When they return to lower states, they emit radiation at specific, discrete frequencies corresponding to the energy difference between quantized orbits, forming the line spectrum.

For Statement II, we use the Rydberg formula for frequency $\nu$: $$\nu = Rc Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

Case 1: Second line of Balmer series for $\text{He}^+$
For Balmer series, $n_1 = 2$. The second line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$.
For $\text{He}^+$, $Z = 2$. $$\nu_{He^+} = Rc (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$$ $$\nu_{He^+} = 4Rc \left( \frac{1}{4} - \frac{1}{16} \right) = 4Rc \left( \frac{4-1}{16} \right) = 4Rc \left( \frac{3}{16} \right) = \frac{3}{4}Rc$$

Case 2: First line of Lyman series for Hydrogen
For Lyman series, $n_1 = 1$. The first line corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.
For $H$, $Z = 1$. $$\nu_{H} = Rc (1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$$ $$\nu_{H} = Rc \left( 1 - \frac{1}{4} \right) = \frac{3}{4}Rc$$

Since $\nu_{He^+} = \nu_{H}$, Statement II is True.

Related Theory

1. Atomic Emission Spectra and the Hydrogen Atom

The study of atomic spectra provided the first evidence that energy in atoms is quantized. When an element in the gaseous state is heated or subjected to an electric discharge, it emits light. If this light is passed through a prism, it does not produce a continuous rainbow (like white light) but rather a series of sharp, isolated lines. This is known as a line emission spectrum. Each element has a unique "fingerprint" spectrum.

In the case of Hydrogen, the simplest atom, the spectrum consists of several series of lines named after their discoverers: Lyman, Balmer, Paschen, Brackett, and Pfund. The discrete nature of these lines was inexplicable by classical physics, which predicted that an electron orbiting a nucleus should lose energy continuously and eventually spiral into the nucleus.

2. The Bohr Model and Quantization

Niels Bohr solved this by postulating that electrons revolve in stationary orbits of fixed energy. Energy is only emitted or absorbed when an electron "jumps" from one orbit to another. The energy of an electron in the $n^{th}$ orbit is given by: $$E_n = -2.18 \times 10^{-18} \left( \frac{Z^2}{n^2} \right) \text{ Joules}$$ This equation shows that as $n$ increases, the energy becomes less negative (higher energy). The negative sign indicates that the electron is bound to the nucleus.

3. The Rydberg Equation

Johannes Rydberg generalized the observations of Balmer into a mathematical formula that predicts the wavelength ($\lambda$) or wave number ($\bar{\nu}$) of any line in the hydrogen spectrum: $$\bar{\nu} = \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ Where $R_H$ is the Rydberg constant ($\approx 1.09677 \times 10^7 \text{ m}^{-1}$). To find frequency ($\nu$), we use $\nu = c / \lambda$.

4. Spectral Series Overview

Series	$n_1$ (Lower)	$n_2$ (Higher)	Spectral Region
Lyman	1	2, 3, 4...	Ultraviolet
Balmer	2	3, 4, 5...	Visible
Paschen	3	4, 5, 6...	Infrared
Brackett	4	5, 6, 7...	Infrared

5. Comparing H-like Species

Bohr's theory applies to "Hydrogen-like" species—ions containing only one electron, such as $\text{He}^+, \text{Li}^{2+}, \text{Be}^{3+}$. The key difference is the atomic number $Z$. Since the frequency is proportional to $Z^2$, an ion with $Z=2$ will have frequencies four times higher than hydrogen for the same transition transitions.

6. Short-cut: Comparing Transitions

To find if two transitions in different species have the same frequency, check if the value of $Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ is identical. For example, a transition from $n=4 \to n=2$ in $\text{He}^+$ ($Z=2$) yields: $$2^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4 \left( \frac{1}{4} - \frac{1}{16} \right) = 1 - \frac{1}{4} = \frac{3}{4}$$ This matches the $n=2 \to n=1$ transition in Hydrogen ($Z=1$): $$1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4}$$

7. Common Pitfalls for Students

Confusing "Line Number" with Orbit: The "first line" of any series always comes from the orbit immediately above the base orbit. For Balmer ($n_1=2$), the first line is $3 \to 2$, the second is $4 \to 2$.
Ignoring Z: Many students forget to square the atomic number ($Z^2$) when dealing with ions like $\text{He}^+$ or $\text{Li}^{2+}$.
Absorption vs Emission: Emission occurs from high to low energy levels. Absorption occurs from low to high. The Lyman series in emission is usually only seen in discharge tubes because, under standard conditions, electrons are in the ground state ($n=1$).

8. Quantum Mechanical Perspective

While Bohr's model explains the line spectrum, it fails to explain the relative intensities of the lines or the "fine structure" (splitting of lines). This required the development of wave mechanics (Schrödinger equation), which replaces fixed orbits with "orbitals"—regions of probability. However, for calculating frequencies of 1-electron systems, Bohr's results remain perfectly accurate.

9. Calculation of Limiting Lines

The "limiting line" or "series limit" of any series occurs when the electron transitions from $n_2 = \infty$ to the base orbit $n_1$. This radiation represents the ionization energy if starting from the ground state. For the Lyman series limit: $$\bar{\nu}_{limit} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty} \right) = R_H$$

10. Exam Relevance

This topic is a cornerstone of Physics and Chemistry in JEE and NEET. Questions often involve: 1. Calculating the number of spectral lines emitted when an electron drops from a high orbit to ground state: $\text{Lines} = \frac{n(n-1)}{2}$. 2. Comparing wavelengths and frequencies across different H-like ions. 3. Identifying the region of the electromagnetic spectrum (UV, Visible, IR).

Frequently Asked Questions (FAQs)

1. Why do we see discrete frequencies instead of a continuous spectrum? Electrons in an atom can only exist in specific, quantized energy levels. Since energy is conserved, the photon emitted during a transition must exactly match the difference between these levels, leading to specific (discrete) frequencies.

2. What is the Rydberg constant value? For calculation purposes, $R \approx 1.097 \times 10^7 \text{ m}^{-1}$. In energy terms, $Rhc \approx 13.6 \text{ eV}$.

3. Which series of the hydrogen spectrum lies in the visible region? The Balmer series is the only one that falls within the visible part of the electromagnetic spectrum.

4. Does Bohr's model work for He gas? No, it only works for 1-electron systems. It works for $\text{He}^+$ ion, but not neutral Helium ($2$ electrons) due to inter-electronic repulsions.

5. What is the 'second line' of a series? It is the transition from $n_1+2$ to $n_1$. For example, in Lyman ($n_1=1$), it is $3 \to 1$.

6. How is frequency related to wavelength? They are inversely proportional: $\nu = c/\lambda$. Shorter wavelength means higher frequency and energy.

7. What happens when electric discharge is passed? Energy is provided to the $H_2$ molecules, causing them to vibrate violently until the bond breaks (dissociation) and the resulting atoms get excited.

8. What is the H-alpha line? It is the first line of the Balmer series ($3 \to 2$), which produces a characteristic red glow.

9. Why is the Lyman series in the UV region? Because the energy gaps between $n=1$ and higher levels are much larger than gaps between other levels, resulting in high-frequency UV radiation.

10. Can an electron transition from n=1 to n=4? Yes, but that would be an absorption transition requiring the atom to take in a photon of specific frequency.

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