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MHT CET · Physics · Atomic Physics

The ratio of longest to shortest wavelength emitted in Paschen series of hydrogen atom is

  1. A \(\frac{144}{63}\)
  2. B \(\frac{25}{9}\)
  3. C \(\frac{9}{25}\)
  4. D \(\frac{63}{144}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{144}{63}\)

Step-by-step Solution

Detailed explanation

For Paschen series, Longest wavelength corresponds to
\(\begin{aligned}
& \mathrm{n}_1=3, \mathrm{n}_2=\mathrm{n}_1+1=4 \\
& \lambda_{\max }=\frac{144}{7 \mathrm{R}}
\end{aligned}\)
Shortest wavelength corresponds to \(\mathrm{n}_1=3, \mathrm{n}_2=\infty\)
\(\begin{aligned}
\lambda_{\min } & =\frac{9}{R} \\
\therefore \quad & \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{\left(\frac{144}{7 R}\right)}{\left(\frac{9}{R}\right)}=\frac{144}{63}
\end{aligned}\)
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