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Q.
If the first line in the Lyman series has wavelength $\lambda$, then the first line in Balmer series has the wavelength
TS EAMCET 2020
Solution:
Wavelength for Lyman series is calculated as
$\frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)$
For first line of Lyman series, $n=2$
$\therefore \frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$
$\Rightarrow \frac{1}{\lambda}=\frac{3 R}{4}$
$\Rightarrow \lambda=\frac{4}{3 R}$
$\Rightarrow R=\frac{4}{3 \lambda}$
Wavelength of Balmer series is calculated as
$\frac{1}{\lambda'}=R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right)$
For first line of Balmer series, $n=3$
$\therefore \frac{1}{\lambda'}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$
$\Rightarrow \frac{1}{\lambda^{\prime}}=\frac{5}{36} R$
$\Rightarrow \frac{1}{\lambda^{\prime}}=\frac{5}{36} \times \frac{4}{3 \lambda}$ [from Eq. (i)]
$\Rightarrow \frac{1}{\lambda'}=\frac{5}{27 \lambda}$
$\Rightarrow \lambda^{\prime}=\frac{27 \lambda}{5}$