Physics - Modern Physics Question with Solution | TestHub

Using Rydberg's formula,

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{{n^2}_1}-\frac{1}{{n^2}_2}\right)$

For the lowest wavelength in Lyman series, $n_1=1, n_2=\infty$.

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right)=R{Z}^2$

For Balmer series, $n_1=2, n_2=\infty$

$\frac{1}{\lambda \text{'}}=R{Z}^2\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right)=\frac{R{Z}^2}{4}$

So,

 $$\begin{gathered} \frac{1}{\lambda \text{'}}=\frac{1}{4\lambda } \\ \Rightarrow \lambda \text{'}=4\times 917 \stackrel{}{ \mathrm{\AA }=3668 \mathrm{\AA }} \end{gathered}$$