Question

If the 9546 \(\AA\) wavelength spectral line in hydrogen spectrum is due to the transition of an electron from a higher orbit to \(n^{th}\) lower orbit, then the value of \(n\) is:

• A. 2
• B. 1
• C. 3
• D. 4

Hint:

Series of hydrogen spectrum: \[ n=1 \rightarrow \text{Lyman} \] \[ n=2 \rightarrow \text{Balmer} \] \[ n=3 \rightarrow \text{Paschen} \] \[ n=4 \rightarrow \text{Brackett} \]

Answer 1

The Correct Option is C

Concept: Rydberg formula: \[ \frac1\lambda = R \left( \frac1{n^2} - \frac1{m^2} \right) \] where \(m>n\).

Step 1: Convert wavelength into metre.
\[ \lambda = 9546\times10^{-10}m. \] \[ = 9.546\times10^{-7}m. \]

Step 2: Calculate the spectral series.
\[ \frac1{\lambda R} = \frac1{(9.546\times10^{-7})(1.097\times10^7)} \] \[ \approx0.095. \] For \[ n=3 \] (Paschen series) \[ \frac1{3^2} = 0.111. \] Difference with \(m=8\): \[ \frac19-\frac1{64} = 0.095. \] This exactly matches the given wavelength.

Step 3: Identify the lower orbit.
Hence transition belongs to \[ n=3. \] Therefore, \[ \boxed{3} \]