Thank you for reporting, we will resolve it shortly
Q.
The ratio of minimum wavelengths of Lyman and Balmer series will be
NTA AbhyasNTA Abhyas 2022
Solution:
The series end of Lyman series corresponds to transition from $n_{ i }=\infty$ to $n_{ f }=1$, corresponding to the wavelength,
$\frac{1}{\left(\lambda_{\min }\right)_{ L }}=R\left[\frac{1}{1}-\frac{1}{\infty}\right]=R$
$\Rightarrow\left(\lambda_{\min }\right)_{ L }=\frac{1}{R}=91.2 \,nm$
For last line of Balmer series,
$\frac{1}{\left(\lambda_{\min }\right)_{B}}=R\left[\frac{1}{(2)^{2}}-\frac{1}{(\infty)^{2}}\right]=\frac{R}{4} $
$\Rightarrow\left(\lambda_{\min }\right)_{B}=\frac{4}{R}=364.8 \,nm$
Dividing equation (i) by (ii), we get,
$\therefore \frac{\left(\lambda_{\min }\right)_{ L }}{\left(\lambda_{\min }\right)_{ B }}=0.25$