Question 106 of 239

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

A$4: 1$B$1: 4$C$2: 1$D$1: 2$

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Answer: (A) $4: 1$

$\begin{aligned} & \frac{1}{\lambda}=\mathrm{Rz}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\ & \frac{\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{Rz}^2\left(\frac{1}{1^2}\right)}{\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{Rz}^2\left(\frac{1}{2^2}\right)} \\ & \frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{L}}}=4: 1\end{aligned}$

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