MHT CET · Chemistry · Structure of Atom
What is the wave number of lowest transition in Balmer series?
- A \(\mathrm{R}_{\mathrm{H}}\left(\frac{36}{5}\right)\)
- B \(\mathrm{R}_{\mathrm{H}}\left(\frac{5}{36}\right)\)
- C \(\mathrm{R}_{\mathrm{H}}\left(\frac{21}{100}\right)\)
- D \(\mathrm{R}_{\mathrm{H}}\left(\frac{100}{21}\right)\)
Answer & Solution
Correct Answer
(B) \(\mathrm{R}_{\mathrm{H}}\left(\frac{5}{36}\right)\)
Step-by-step Solution
Detailed explanation
The lowest transition in Balmer series is from \(\mathrm{n}_1=2\) and \(\mathrm{n}_2=3\)
\(\therefore \overline{\mathrm{v}}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1{ }^2}-\frac{1}{\mathrm{n}_2^2}\right]=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]\)\(=\mathrm{R}_{\mathrm{H}}\left[\frac{5}{36}\right]\)
\(\therefore \overline{\mathrm{v}}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1{ }^2}-\frac{1}{\mathrm{n}_2^2}\right]=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]\)\(=\mathrm{R}_{\mathrm{H}}\left[\frac{5}{36}\right]\)
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