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TS EAMCET · Physics · Atomic Physics

In hydrogen spectrum, the shortest and longest wavelengths of Balmer series are \(\lambda_1\) and \(\lambda_2\) respectively. The Rydberg constant of hydrogen is

  1. A \(\frac{1}{\lambda_1}-\frac{9}{\lambda_2}\)
  2. B \(\frac{4}{\lambda_1}-\frac{9}{\lambda_2}\)
  3. C \(\frac{9}{\lambda_1}-\frac{9}{\lambda_2}\)
  4. D \(\frac{9}{\lambda_1}-\frac{4}{\lambda_2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{9}{\lambda_1}-\frac{9}{\lambda_2}\)

Step-by-step Solution

Detailed explanation

For the stortest wavelength \(\lambda\); of Balmer series, \(\mathrm{n}_1=\) 2 and \(n_2=\infty\) \(\begin{aligned} & \frac{1}{\lambda_1}=\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) . \\ & \frac{1}{\lambda_1}=\frac{\mathrm{R}}{4} ...(1) \end{aligned}\) For the…
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