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

• Correct answer
  A

  $4: 1$
• B

  $1: 4$
• C

  $2: 1$
• D

  $1: 2$

Solution

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

Hence, the correct answer is option: 'A'.