The first line in the Lyman series has wavele | vedclass

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Question

For first line in Lyman series $\lambda_{L_{1}}=\frac{4}{3 R}$           ...........$(i)$

For first line in Balmer se

Vedclass · Accepted Answer

$\frac{27}{5}\lambda $

Vedclass · Answer

For first line in Lyman series $\lambda_{L_{1}}=\frac{4}{3 R}$           ...........$(i)$

For first line in Balmer series $\lambda_{B_{1}}=\frac{36}{5 R}$            ............$(ii)$

From equations $(i)$ and $(ii)$

$\frac{\lambda_{\mathrm{B}_{1}}}{\lambda_{\mathrm{L}_{1}}}=\frac{27}{5}$

$ \Rightarrow \lambda_{\mathrm{B}_{1}} \frac{27}{5} \lambda_{\mathrm{L}_{1}} $

$\Rightarrow \lambda_{\mathrm{B}_{1}} \frac{27}{5} \lambda$

The first line in the Lyman series has wavelength $\lambda $. The wavlength of the first line in Balmer series is

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