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: The user's provided answer sheet indicates option (A) was chosen. However, a detailed analysis shows that both statements are true based on the standard Bohr model, which is the expected level for this exam. The reasoning for Statement II being false is extremely subtle and usually ignored.)*
Step 1: Understanding the Question:
We need to evaluate two statements related to the atomic structure and spectra of hydrogen and hydrogen-like ions.
Step 2: Detailed Explanation:
Analysis of Statement I:
\begin{itemize} \item The statement describes the process of generating an atomic emission spectrum for hydrogen.
\item Passing an electric discharge provides energy to the H₂ gas.
\item This energy first causes the dissociation of H₂ molecules into individual H atoms: \( H_2 \rightarrow 2H \).
\item The H atoms then absorb energy, causing their electrons to get excited to higher energy levels.
\item These excited electrons are unstable and fall back to lower energy levels, emitting the excess energy as photons of electromagnetic radiation.
\item According to Bohr's model, the energy levels in an atom are quantized (discrete). Therefore, the energy difference between any two levels is fixed, leading to the emission of photons with specific, discrete frequencies (or wavelengths). This creates a line spectrum.
\item Conclusion: Statement I is a correct description of the phenomenon. Statement I is true.
\end{itemize} Analysis of Statement II:
\begin{itemize} \item We can use the Rydberg formula for the frequency (\(\nu\)) of spectral lines:
\[ \nu = c \cdot \bar{\nu} = c \cdot R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \(R_H\) is the Rydberg constant, \(c\) is the speed of light, and \(Z\) is the atomic number.
\item For He⁺, second line of Balmer series: \begin{itemize} \item Balmer series means the final level is \(n_1 = 2\).
\item The second line corresponds to a transition from \(n_2 = 4\).
\item For Helium (He⁺), the atomic number is \(Z = 2\).
\item \[ \nu_{He^+} = c \cdot R_H \cdot (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = c \cdot R_H \cdot 4 \left( \frac{1}{4} - \frac{1}{16} \right) = c \cdot R_H \cdot 4 \left( \frac{4-1}{16} \right) = c \cdot R_H \cdot 4 \left( \frac{3}{16} \right) = \frac{3}{4} c R_H \] \end{itemize} \item For H, first line of Lyman series: \begin{itemize} \item Lyman series means the final level is \(n_1 = 1\).
\item The first line corresponds to a transition from \(n_2 = 2\).
\item For Hydrogen (H), the atomic number is \(Z = 1\).
\item \[ \nu_{H} = c \cdot R_H \cdot (1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = c \cdot R_H \cdot 1 \left( 1 - \frac{1}{4} \right) = c \cdot R_H \left( \frac{3}{4} \right) = \frac{3}{4} c R_H \] \end{itemize} \item Comparison: The calculated frequencies are equal: \( \nu_{He^+} = \nu_{H} \).
\item Conclusion: Statement II is correct based on the Bohr model formula. Statement II is true.
\end{itemize} Step 3: Final Answer:
Both Statement I and Statement II are true. Therefore, the correct option is (B).