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AP EAMCET · PHYSICS · Atomic Physics

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

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

Answer & Solution

Correct Answer

(B) \(\frac{4\left(\lambda_2-\lambda_1\right)}{3 \lambda_1 \lambda_2}\)

Step-by-step Solution

Detailed explanation

For Lymen series \(\begin{aligned} & \frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \mathrm{n}_1=1 \text { and } \mathrm{n}_2=\infty\end{aligned}\) \(\frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}}\) ...(1) For balmer…