Physics - Modern Physics Question with Solution | TestHub

Using Rydberg's formula,

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

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, $n_1=2,n_2=\infty$

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

So,

$$\begin{gathered} \frac{1}{{\lambda }^{\prime }}=\frac{1}{4\lambda } \\ \Rightarrow {\lambda }^{\prime }=4\times 917\mathring{A}=3668\mathring{A} \end{gathered}$$