Question:
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?
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?
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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 \(r_1 \approx 0.53 \text{ \AA}\), \(9 \times 0.53 \approx 4.77 \text{ \AA}\).
Updated On: Mar 26, 2026
- Lyman
- Paschen
- Balmer
- Pfund
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The Correct Option is B
Solution and Explanation
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: Calculate 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:
Paschen series.
Step 2: Calculate 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:
- Lyman series: \(n_f = 1\)
- Balmer series: \(n_f = 2\)
- Paschen series: \(n_f = 3\)
- Brackett series: \(n_f = 4\)
- Pfund series: \(n_f = 5\)
Paschen series.
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