Physics - Modern Physics Question with Solution | TestHub

Here, for wavelength ${\lambda }_1$ $n_1=4$ and $n_2=3$ and for ${\lambda }_2,n_1=3$ and $n_2=2$ We have $\frac{hc}{\lambda }=-13.6\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]$ So, for ${\lambda }_1$ $\begin{array}{rl}\Rightarrow \frac{hc}{{\lambda }_1}& =-13.6\left[\frac{1}{(4{)}^2}-\frac{1}{(3{)}^2}\right]\\ \frac{hc}{{\lambda }_1}& =13.6\left[\frac{7}{144}\right]\end{array}$ Similarly, for ${\lambda }_2$ $\begin{array}{rl}\Rightarrow \frac{hc}{{\lambda }_2}& =-13.6\left[\frac{1}{(3{)}^2}-\frac{1}{(2{)}^2}\right]\\ \frac{hc}{{\lambda }_2}& =13.6\left[\frac{5}{36}\right]\end{array}$ Hence, from Eqs. (i) and (ii), we get $\frac{{\lambda }_1}{{\lambda }_2}=\frac{20}{7}$