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

Step 1: Know what controls the wavelength.
For any atom that has only one electron (like H or $He^+$), the line wavelength comes from the Rydberg formula $\frac{1}{\lambda}=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$. Here $Z$ is the nuclear charge.
Step 2: See which parts stay the same.
The question says it is the SAME line of the Balmer series. So the levels $n_1$ and $n_2$ are the same for both atoms. That means the bracket part and $R$ do not change.
Step 3: Keep only what changes.
Only $Z$ is different between H and $He^+$. So $\frac{1}{\lambda}\propto Z^2$, which flips to $\lambda\propto \frac{1}{Z^2}$. A bigger charge pulls the electron harder and gives a shorter wave.
Step 4: Put in the charges.
Hydrogen has $Z=1$ and the helium ion $He^+$ has $Z=2$.
Step 5: Make a simple ratio.
\[ \frac{\lambda_{He^+}}{\lambda_{H}}=\frac{Z_H^2}{Z_{He}^2}=\frac{1^2}{2^2}=\frac{1}{4} \] So the helium line is one fourth as long.
Step 6: Find the number.
\[ \lambda_{He^+}=\frac{656.4}{4}=164.1\ \text{nm} \] \[ \boxed{164.1\ \text{nm}} \]