Question:
The shortest wavelength in the Paschen series of the Hydrogen spectrum is
\[
\text{Rydberg constant of hydrogen }=1.097\times 10^7\ \text{m}^{-1}
\]
The shortest wavelength in the Paschen series of the Hydrogen spectrum is
\[ \text{Rydberg constant of hydrogen }=1.097\times 10^7\ \text{m}^{-1} \]
\[ \text{Rydberg constant of hydrogen }=1.097\times 10^7\ \text{m}^{-1} \]
Show Hint
For the shortest wavelength in any hydrogen spectral series, use the series limit:
\[
n_2\to \infty
\]
For Paschen series:
\[
n_1=3
\]
Always substitute the lower energy level correctly.
Updated On: Jun 25, 2026
- \(91.2\ \text{nm}\)
- \(364.6\ \text{nm}\)
- \(820.4\ \text{nm}\)
- \(2278.9\ \text{nm}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Use the Rydberg formula for Hydrogen spectrum.
The wavelength of spectral lines in hydrogen is given by \[ \frac{1}{\lambda} = R\left( \frac{1}{n_1^2}-\frac{1}{n_2^2} \right) \] where \[ R=1.097\times 10^7\ \text{m}^{-1} \] For the Paschen series, \[ n_1=3 \] The shortest wavelength corresponds to the series limit: \[ n_2\to \infty \]
Step 2: Apply the series limit condition.
Since \[ \frac{1}{\infty^2}=0, \] the formula becomes \[ \frac{1}{\lambda_{\min}} = R\left(\frac{1}{3^2}\right) \] \[ \frac{1}{\lambda_{\min}} = \frac{R}{9} \] Thus, \[ \lambda_{\min}=\frac{9}{R} \]
Step 3: Substitute the value of Rydberg constant.
\[ \lambda_{\min} = \frac{9}{1.097\times 10^7} \] \[ \lambda_{\min} = 8.204\times 10^{-7}\ \text{m} \]
Step 4: Convert into nanometer.
Since \[ 1\ \text{nm}=10^{-9}\ \text{m}, \] we get \[ \lambda_{\min} = 8.204\times 10^{-7}\times 10^9\ \text{nm} \] \[ \lambda_{\min}=820.4\ \text{nm} \]
Step 5: Final conclusion.
Hence, the shortest wavelength in the Paschen series is \[ \boxed{820.4\ \text{nm}} \]
The wavelength of spectral lines in hydrogen is given by \[ \frac{1}{\lambda} = R\left( \frac{1}{n_1^2}-\frac{1}{n_2^2} \right) \] where \[ R=1.097\times 10^7\ \text{m}^{-1} \] For the Paschen series, \[ n_1=3 \] The shortest wavelength corresponds to the series limit: \[ n_2\to \infty \]
Step 2: Apply the series limit condition.
Since \[ \frac{1}{\infty^2}=0, \] the formula becomes \[ \frac{1}{\lambda_{\min}} = R\left(\frac{1}{3^2}\right) \] \[ \frac{1}{\lambda_{\min}} = \frac{R}{9} \] Thus, \[ \lambda_{\min}=\frac{9}{R} \]
Step 3: Substitute the value of Rydberg constant.
\[ \lambda_{\min} = \frac{9}{1.097\times 10^7} \] \[ \lambda_{\min} = 8.204\times 10^{-7}\ \text{m} \]
Step 4: Convert into nanometer.
Since \[ 1\ \text{nm}=10^{-9}\ \text{m}, \] we get \[ \lambda_{\min} = 8.204\times 10^{-7}\times 10^9\ \text{nm} \] \[ \lambda_{\min}=820.4\ \text{nm} \]
Step 5: Final conclusion.
Hence, the shortest wavelength in the Paschen series is \[ \boxed{820.4\ \text{nm}} \]
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