If Rydberg's constant is $R$, the longest wavelength of radiation in Paschen series will be $\frac{\alpha}{7 R}$, where $\alpha=$ ________.
Answer (integer)
144
Solution
<p>Longest wavelength corresponds to transition between $\mathrm{n}=3$ and $\mathrm{n}=4$</p>
<p>$$\begin{aligned}
& \frac{1}{\lambda}=\mathrm{RZ}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\mathrm{RZ}^2\left(\frac{1}{9}-\frac{1}{16}\right) \\
& =\frac{7 \mathrm{RZ}^2}{9 \times 16} \\
& \Rightarrow \lambda=\frac{144}{7 \mathrm{R}} \text { for } \mathrm{Z}=1 \quad \therefore \alpha=144
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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