Question

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 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. Statement I is true but Statement II is false
• B. Both Statement I and Statement II are true
• C. Statement I is false but Statement II is true
• D. Both Statement I and Statement II are false

Hint:

For hydrogen-like species (one electron), the energy of a level is proportional to \(Z^2/n^2\). This means a transition in a species with atomic number Z from \(n_{2,Z}\) to \(n_{1,Z}\) will have the same energy/frequency as a transition in hydrogen (Z=1) from \(n_{2,H}\) to \(n_{1,H}\) if \(Z/n_{Z} = 1/n_{H}\). In this case, for He⁺ (Z=2), the transition 4 → 2 is equivalent to hydrogen's 2 → 1 transition because 2/4 = 1/2 and 2/2 = 1/1.

Answer 1

The Correct Option is B

(Note: Although the provided answer key marks option (A), a standard Bohr-model–based analysis at the expected exam level shows that both statements are true. The subtle refinement sometimes used to reject Statement II is beyond the intended scope of this question.)

Step 1: Understanding the Question:

The question asks us to evaluate two statements related to the atomic structure of hydrogen and hydrogen-like ions, and the origin of their emission spectra.

Step 2: Detailed Explanation:

Analysis of Statement I:

The statement describes the formation of the atomic emission spectrum of hydrogen.

When an electric discharge is passed through hydrogen gas, energy is supplied to the system. This energy first causes dissociation of hydrogen molecules into atoms:
\( H_2 \rightarrow 2H \)

The hydrogen atoms then absorb energy, exciting their electrons to higher energy levels. These excited states are unstable, and electrons return to lower energy levels, emitting excess energy as photons.

According to the Bohr model, the energy levels of an atom are quantized. Therefore, transitions between fixed energy levels result in emission of photons with specific energies (frequencies or wavelengths), producing a **line spectrum**.

Conclusion:
Statement I correctly explains the origin of the hydrogen emission spectrum. Statement I is true.

Analysis of Statement II:

The frequency of spectral lines for hydrogen-like species is given by the Rydberg relation:

\[ \nu = c R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For He⁺ (second line of Balmer series):

Balmer series corresponds to transitions ending at \( n_1 = 2 \). The second line corresponds to \( n_2 = 4 \). For He⁺, the atomic number \( Z = 2 \).

\[ \nu_{He^+} = c R_H (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = c R_H \cdot 4 \left( \frac{1}{4} - \frac{1}{16} \right) = c R_H \cdot 4 \left( \frac{3}{16} \right) = \frac{3}{4} c R_H \]

For H (first line of Lyman series):

Lyman series corresponds to transitions ending at \( n_1 = 1 \). The first line corresponds to \( n_2 = 2 \). For hydrogen, \( Z = 1 \).

\[ \nu_H = c R_H (1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = c R_H \left( 1 - \frac{1}{4} \right) = \frac{3}{4} c R_H \]

Comparison:
\[ \nu_{He^+} = \nu_H \]

Conclusion:
Based on the Bohr model, the frequencies are equal. Statement II is true.

Step 3: Final Answer:

Both Statement I and Statement II are true. Correct option: (B)