Question

Sol. [Key Concept: Lyman shortest: $\frac{1}{x} = R{(1)}^{2}(1 - 0)$; Balmer longest of He$+$ ($Z = 2$): $\frac{1}{\lambda} = R{(2)}^{2}(\frac{1}{4} - \frac{1}{9})$; divide to get $\lambda = \frac{9x}{5}$.]

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

$$\begin{array}{r} \frac{1}{\lambda} = R \times 2^{2} \times \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)\#(2) \end{array}$$

$${\frac{x}{\lambda} = R \times 4 \times \frac{5}{4 \times 9} }{\frac{1}{\lambda} = \frac{1}{x} \times \frac{5}{9} }{\lambda = \frac{9x}{5} }$$