Question:
The shortest wavelength in the Balmer series of hydrogen atom spectrum is [Rydberg constant \(R = 1.097 \times 10^{7}\, m^{-1}\)]
The shortest wavelength in the Balmer series of hydrogen atom spectrum is [Rydberg constant \(R = 1.097 \times 10^{7}\, m^{-1}\)]
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Shortest wavelength in any series corresponds to transition from infinity to the final level.
Updated On: Jul 18, 2026
- 91.2 nm
- 364.6 nm
- 820.4 nm
- 2278.9 nm
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The Correct Option is B
Solution and Explanation
Step 1: Understand Balmer series condition.
In Balmer series, transitions occur from higher levels to \(n_1 = 2\). The shortest wavelength corresponds to maximum energy transition.
Step 2: Identify limiting case.
Shortest wavelength occurs when: \[ n_2 \to \infty \] So we use Rydberg formula: \[ \frac{1}{\lambda} = R\left(\frac{1}{2^2} - 0\right) \]
Step 3: Apply formula.
\[ \frac{1}{\lambda} = \frac{R}{4} \]
Step 4: Substitute value of R.
\[ \lambda = \frac{4}{1.097 \times 10^7} \]
Step 5: Compute wavelength.
\[ \lambda \approx 3.646 \times 10^{-7}\, m \]
Step 6: Convert to nm.
\[ \lambda = 364.6\, nm \]
Final Answer:
\[ \boxed{364.6\, nm} \]
In Balmer series, transitions occur from higher levels to \(n_1 = 2\). The shortest wavelength corresponds to maximum energy transition.
Step 2: Identify limiting case.
Shortest wavelength occurs when: \[ n_2 \to \infty \] So we use Rydberg formula: \[ \frac{1}{\lambda} = R\left(\frac{1}{2^2} - 0\right) \]
Step 3: Apply formula.
\[ \frac{1}{\lambda} = \frac{R}{4} \]
Step 4: Substitute value of R.
\[ \lambda = \frac{4}{1.097 \times 10^7} \]
Step 5: Compute wavelength.
\[ \lambda \approx 3.646 \times 10^{-7}\, m \]
Step 6: Convert to nm.
\[ \lambda = 364.6\, nm \]
Final Answer:
\[ \boxed{364.6\, nm} \]
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