Question

Given below are two statements:
Statement I: When electric discharge is put on hydrogen, it emits discrete frequency in the electromagnetic spectrum.
Statement II: Frequency of He\(^+\) ion of 2\(^\text{nd}\) line of Balmer series is equal to first line of Lyman series.
In the light of the above statements, choose the correct option.

• A. Both statement I and statement II are correct.
• B. Both statement I and statement II are incorrect.
• C. Statement I is correct and statement II is incorrect.
• D. Statement I is incorrect and statement II is correct.

Hint:

The frequency of spectral lines can be derived from the Rydberg formula, where the difference in the inverse square of the integers determines the frequency of the emitted radiation.

Answer 1

The Correct Option is A

Let's analyze the two statements based on known physics concepts:

1. Statement I: "When electric discharge is put on hydrogen, it emits discrete frequency in the electromagnetic spectrum."
   • Hydrogen atoms emit light in discrete wavelengths (frequencies) when subjected to an electric discharge due to electron transitions between different energy levels in the atom.
   • This phenomenon is known as the emission spectrum of hydrogen, where each line corresponds to a specific electron transition.
   • The emitted lines form a series, such as the Lyman, Balmer, and Paschen series, each corresponding to electron transitions to a specific lower energy level (n = 1 for Lyman, n = 2 for Balmer, etc.).
   • Since the statement accurately describes the emission of discrete frequencies during electric discharge in hydrogen, Statement I is correct.
2. Statement II: "Frequency of He\(^+\) ion of 2\(^\text{nd}\) line of Balmer series is equal to first line of Lyman series."
   • The Bohr model equation for the frequency of emitted light is given by: \(f = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)Z^2\)
   • For hydrogen (\(Z = 1\)), the 1st line of the Lyman series (n₁ = 1, n₂ = 2): \(f_{\text{Lyman}} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = \frac{3R}{4}\)
   • For He\(^+\) ion (\(Z = 2\)) with the 2nd line of the Balmer series (n₁ = 2, n₂ = 4): \(f_{\text{Balmer}} = R\left(\frac{1}{2^2} - \frac{1}{4^2}\right) \cdot 2^2 = R\left(\frac{3}{16}\right) \cdot 4 = \frac{3R}{4}\)
   • Comparing both frequencies: \(f_{\text{Lyman}} = \frac{3R}{4} = f_{\text{Balmer}}\), thus they are equal.
   • Therefore, Statement II is also correct.

In conclusion, both Statement I and Statement II are correct.