The electron in hydrogen atom undergoes transition from higher orbits to an orbit of radius 476.1 pm. This transition corresponds to which of the following series?
Show Hint
Remember the order of hydrogen spectral series: Lyman (1), Balmer (2), Paschen (3), Brackett (4), Pfund (5). The radius scales with \(n^2\), so \(n=3\) means \(r \approx 9 \times r_1\). Since \(r_1 \approx 0.53 \text{ \AA}\), \(9 \times 0.53 \approx 4.77 \text{ \AA}\).
Step 1: Formula for Radius of Orbit
The radius of the \(n^{\text{th}}\) orbit of a hydrogen atom is given by:
\[ r_n = 0.529 \times \frac{n^2}{Z} \ \text{\AA} \]
For Hydrogen, \(Z = 1\).
Given radius, \(r = 476.1 \text{ pm} = 4.761 \text{ \AA}\).
Step 2: Calculation Process the Orbit Number (\(n\))
Substitute the values into the formula:
\[ 4.761 = 0.529 \times n^2 \]
\[ n^2 = \frac{4.761}{0.529} \approx 9 \]
\[ n = \sqrt{9} = 3 \]
So, the electron transitions to the orbit \(n = 3\).
Step 3: Identify the Spectral Series
Spectral series are defined by the final orbit (\(n_f\)) of the transition:
\begin{itemize}
\item Lyman series: \(n_f = 1\)
\item Balmer series: \(n_f = 2\)
\item Paschen series: \(n_f = 3\)
\item Brackett series: \(n_f = 4\)
\item Pfund series: \(n_f = 5\)
\end{itemize}
Since the transition ends at \(n = 3\), it corresponds to the Paschen series.
Required Answer: Paschen series.