If the wavelength of the n^ th line of Lyman series is equal to the...

n^{\text{th }}$line of Lyman series means transition from$(n+1)$th state to first state.$$\begin{aligned}& \frac{1}{\lambda}=R Z²\left[1-\frac{1}{(n+1)^2}\right]\\ & \text{ de-Broglie wavelength, } \quad \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{(2 \pi)(h r)}{(n+1) h}=\frac{2 \pi r}{(n+1)}\\ & \frac{1}{\lambda}=\frac{(n+1)}{2 \pi r}\\ & \end{aligned}$$or Equating (i) and (ii), we get$\left(\frac{n+1}{2 \pi r}\right)=$R Z²\left[\frac{n(n+2)}{(n+1)^2}\right]$Now, as $$\begin{aligned}& r \propto \frac{n^2}{Z}\\ & r=\frac{(n+1)^2}{11}r_0 \end{aligned}$$\therefore \quad r=$\frac{(n+1)^2}{11}$r_0 Substituting in equations (iii), we get $\frac{11}{2 \pi r_0}=\frac{R(11)^2(n)(n+2)}{(n+1)}$or (n+1)=\left(1.09 $\times$10^7\right)(11)(2 \pi) \times\left(0.529 $\times$10^{-10}\right)\left(n^2+2 n\right) Solving this equation we get,n=24