(a) Sketch the energy level diagram for hydrogen atom. (b) Find the ratio of the longest and the shortest wavelength in Lyman series in hydrogen atom.

$\displaystyle \frac{1}{\lambda}=R\left\lfloor\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right\rfloor$

For longest wavelength in Lyman series

$$n_{$\displaystyle 1$}=$\displaystyle 1$, n_{$\displaystyle 2$}=$\displaystyle 2$

$$$$ \begin{aligned} & \frac{1}{\lambda_{$\displaystyle 1$}}=R\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]=R\left[1-\frac{1}{4}\right]=\frac{3}{4} R & \therefore \lambda_{$\displaystyle 1$}=\frac{4}{3 R} \end{aligned}

$$For shortest wavelength in Lyman Series

$$n_{$\displaystyle 1$}=$\displaystyle 1$, n_{$\displaystyle 2$}=\infty

$$$$ \begin{aligned} & \frac{1}{\lambda_{$\displaystyle 2$}}=R\left[\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right]=R[$\displaystyle 1$-$\displaystyle 0$]=R & \therefore \lambda_{$\displaystyle 2$}=\frac{1}{R} \end{aligned}