MHT CET · Chemistry · Structure of Atom
What is the wave number of lowest transition associated with Lyman series?
- A \(\bar{v}=\mathrm{R}_{\mathrm{H}}\left(\frac{3}{4}\right)\)
- B \(\bar{v}=\mathrm{R}_{\mathrm{H}}\left(\frac{5}{36}\right)\)
- C \(\bar{v}=\mathrm{R}_{\mathrm{H}}\left(\frac{4}{3}\right)\)
- D \(\bar{v}=\mathrm{R}_{\mathrm{H}}\left(\frac{36}{5}\right)\)
Answer & Solution
Correct Answer
(A) \(\bar{v}=\mathrm{R}_{\mathrm{H}}\left(\frac{3}{4}\right)\)
Step-by-step Solution
Detailed explanation
For Lyman series, \(\mathrm{n}_1=1\)
The wave number of lowest transition associated with Lyman series corresponds to the transition from \(n=2\) to \(n=1\).
Therefore, \(\mathrm{n}_2=2\)
\(\begin{aligned}
\therefore \quad \bar{v} & =\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \mathrm{cm}^{-1} \\
& =\mathrm{R}_{\mathrm{H}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \mathrm{cm}^{-1}
\end{aligned}\)
\(=\mathrm{R}_{\mathrm{H}}\left[1-\frac{1}{4}\right] \mathrm{cm}^{-1}=\mathrm{R}_{\mathrm{H}}\left[\frac{3}{4}\right] \mathrm{cm}^{-1}\)
The wave number of lowest transition associated with Lyman series corresponds to the transition from \(n=2\) to \(n=1\).
Therefore, \(\mathrm{n}_2=2\)
\(\begin{aligned}
\therefore \quad \bar{v} & =\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \mathrm{cm}^{-1} \\
& =\mathrm{R}_{\mathrm{H}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \mathrm{cm}^{-1}
\end{aligned}\)
\(=\mathrm{R}_{\mathrm{H}}\left[1-\frac{1}{4}\right] \mathrm{cm}^{-1}=\mathrm{R}_{\mathrm{H}}\left[\frac{3}{4}\right] \mathrm{cm}^{-1}\)
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