Physics - Modern Physics Question with Solution | TestHub

$n^{\text{th }}$ line of Lyman series means transition from $(n+1)$ th state to first state. $$\begin{array}{rl}& \frac{1}{\lambda }=R{Z}^2\left[1-\frac{1}{(n+1{)}^2}\right]\\ & \text{ de-Broglie wavelength, }\lambda =\frac{h}{mv}=\frac{hr}{mvr}=\frac{(2\pi )(hr)}{(n+1)h}=\frac{2\pi r}{(n+1)}\\ & \frac{1}{\lambda }=\frac{(n+1)}{2\pi r}\\ & \end{array}$$ or Equating (i) and (ii), we get $$\left(\frac{n+1}{2\pi r}\right)=R{Z}^2\left[\frac{n(n+2)}{(n+1{)}^2}\right]$$ Now, as $$\begin{array}{rl}& r\propto \frac{n^2}{Z}\\ & r=\frac{(n+1{)}^2}{11}{r}_0\end{array}$$ $$\therefore r=\frac{(n+1{)}^2}{11}{r}_0$$ Substituting in equations (iii), we get $$\frac{11}{2\pi r_0}=\frac{R(11{)}^2(n)(n+2)}{(n+1)}$$ or $$(n+1)=\left(1.09\times 10^7\right)(11)(2\pi )\times \left(0.529\times 10^{-10}\right)\left(n^2+2n\right)$$ Solving this equation we get, $n=24$