Question:
The electron in hydrogen atom undergoes transition from higher orbits to an orbit of radius \(476.1 \text{ pm}\). This transition corresponds to which of the following series?
The electron in hydrogen atom undergoes transition from higher orbits to an orbit of radius \(476.1 \text{ pm}\). This transition corresponds to which of the following series?
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Remember the sequence: Lyman (\(1\)), Balmer (\(2\)), Paschen (\(3\)), Brackett (\(4\)), Pfund (\(5\)). Also, \(52.9 \text{ pm}\) is equivalently remembered as \(0.529 \text{ \AA}\).
Updated On: May 11, 2026
- Lyman
- Paschen
- Balmer
- Pfund
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We are given the radius of the target orbit for an electron transition in a hydrogen atom. We must first determine the principal quantum number (\(n\)) corresponding to this radius and then identify the spectral series associated with transitions ending at this \(n\).
Step 2: Key Formula or Approach:
The radius of the \(n\)-th Bohr orbit in a hydrogen-like atom is given by: \[ r_n = 52.9 \times \frac{n^2}{Z} \text{ pm} \] For a Hydrogen atom, \(Z = 1\), making it \(r_n = 52.9 \times n^2 \text{ pm}\).
Step 3: Detailed Explanation:
We are given that the radius of the final orbit is \(476.1 \text{ pm}\). Substitute this into the formula: \[ 52.9 \times n^2 = 476.1 \] Solving for \(n^2\): \[ n^2 = \frac{476.1}{52.9} \] Removing the decimal point gives \(n^2 = \frac{4761}{529}\). By calculating, we find that \(529 \times 9 = 4761\), so: \[ n^2 = 9 \implies n = 3 \] The electron transitions from a higher orbit down to the orbit \(n = 3\). Transitions ending at different principal quantum numbers are categorized as:
Step 4: Final Answer:
The transition corresponds to the Paschen series.
We are given the radius of the target orbit for an electron transition in a hydrogen atom. We must first determine the principal quantum number (\(n\)) corresponding to this radius and then identify the spectral series associated with transitions ending at this \(n\).
Step 2: Key Formula or Approach:
The radius of the \(n\)-th Bohr orbit in a hydrogen-like atom is given by: \[ r_n = 52.9 \times \frac{n^2}{Z} \text{ pm} \] For a Hydrogen atom, \(Z = 1\), making it \(r_n = 52.9 \times n^2 \text{ pm}\).
Step 3: Detailed Explanation:
We are given that the radius of the final orbit is \(476.1 \text{ pm}\). Substitute this into the formula: \[ 52.9 \times n^2 = 476.1 \] Solving for \(n^2\): \[ n^2 = \frac{476.1}{52.9} \] Removing the decimal point gives \(n^2 = \frac{4761}{529}\). By calculating, we find that \(529 \times 9 = 4761\), so: \[ n^2 = 9 \implies n = 3 \] The electron transitions from a higher orbit down to the orbit \(n = 3\). Transitions ending at different principal quantum numbers are categorized as:
- \(n=1\): Lyman Series
- \(n=2\): Balmer Series
- \(n=3\): Paschen Series
- \(n=4\): Brackett Series
- \(n=5\): Pfund Series
Step 4: Final Answer:
The transition corresponds to the Paschen series.
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