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The first line in the Lyman series has wavelength $\lambda $. The wavlength of the first line in Balmer series is
A
$\frac{2}{9}\lambda $
B
$\frac{9}{2}\lambda $
C
$\frac{5}{27}\lambda $
D
$\frac{27}{5}\lambda $
Solution
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}} $
$\Rightarrow \lambda_{\mathrm{B}_{1}} \frac{27}{5} \lambda$
Standard 12
Physics
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