If Rydberg’s constant is \( R \), the longest wavelength of radiation in Paschen series will be \( \frac{\alpha}{7R} \), where \( \alpha = \_\_\_\_\_\_\_ \).
Correct Answer: 144
Approach Solution - 1
To determine the value of \( \alpha \) for the longest wavelength of radiation in the Paschen series expressed as \( \frac{\alpha}{7R} \), we need to analyze the Paschen series of hydrogen. The series is defined for transitions from higher energy levels to \( n=3 \). The wavelength of emitted radiation is given by the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] For the longest wavelength, the smallest possible \( n_2 \) above 3 is 4 (\( n_1=3, n_2=4 \)): \[ \frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \] Simplifying gives: \[ \frac{1}{9} - \frac{1}{16} = \frac{16-9}{144} = \frac{7}{144} \] Thus, \[ \frac{1}{\lambda} = \frac{7R}{144} \] Therefore, \[ \lambda = \frac{144}{7R} \] Comparing this with \( \frac{\alpha}{7R} \), it is clear that \( \alpha = 144 \), confirming the value falls within the expected range of 144. The longest wavelength of the Paschen series is \( \frac{144}{7R} \), validating \( \alpha = 144 \).
Approach Solution -2
The Paschen series corresponds to transitions to \(n = 3\). The longest wavelength corresponds to the transition between \(n = 4\) and \(n = 3\). The inverse wavelength is given by:
\(\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)
For \(n_1 = 3\) and \(n_2 = 4\), and taking \(Z = 1\):
\(\frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right)\)
\(\frac{1}{\lambda} = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144}\)
Thus:
\(\alpha = 144\)
The Correct answer is: 144
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