Question

Wavelength of a particular line in Balmer series of atomic spectrum of hydrogen is 656.4 nm. What is the wavelength (in nm) of corresponding line in the spectrum of $He^{+1}$?

• A. 328.2
• B. 164.1
• C. 492.3
• D. 246.1

Hint:

For hydrogen-like species undergoing the same electronic transition as a hydrogen spectral line, the wavelength is always divided by $Z^2$. This relationship is a direct consequence of the Rydberg equation and provides a shortcut for comparing different one-electron systems.

Answer 1

The Correct Option is B

Concept: The wavelength ($\lambda$) of a spectral line for a hydrogen-like species is governed by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For the same electronic transition, the wavelength is inversely proportional to the square of the atomic number ($Z^2$), meaning $\lambda \propto \frac{1}{Z^2}$.

Step 1: Identify the atomic numbers (Z) for Hydrogen and Helium ion.
For Hydrogen ($H$), the atomic number $Z = 1$. For the Helium ion ($He^+$), the atomic number $Z = 2$.

Step 2: Establish the relationship between wavelengths.
Since $\lambda_{ion} = \frac{\lambda_H}{Z^2}$, we apply this to the helium ion: \[ \lambda_{He^+} = \frac{\lambda_H}{Z_{He}^2} = \frac{\lambda_H}{2^2} = \frac{\lambda_H}{4} \]

Step 3: Calculate the final wavelength.
Given the initial wavelength for hydrogen $\lambda_H = 656.4$ nm, we calculate the wavelength for the $He^+$ ion as follows: \[ \lambda_{He^+} = \frac{656.4 \text{ nm}}{4} = 164.1 \text{ nm} \]