An electron in hydrogen like atom makes a transition from n^th orbit and emits radiation corresponding to Lyman series. If de-Broglie wavelength of electron in n^th orbit is equal to the wavelength of radiation emitted, find the value of n. The atomic number of atom is 11.

If $\lambda$ is the de-Broglie wavelength, then for n^th orbit $2r_n=n\lambda$
Where $r_n=\frac{{\in }_0{h}^2{n}^2}{\pi m{e}^{2}Z}$
$\therefore \frac{1}{\lambda }=\frac{m{e}^{2}Z}{2{\in }_0{h}^{2}n}\mathrm{......}\left(i\right)$
For Lyman series $\frac{1}{\lambda }=Z^{2}R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\mathrm{.......}\left(ii\right)$
From equations (i) and (ii) $Z^{2}R\left(1-\frac{1}{n^2}\right)=\frac{m{e}^{2}Z}{2{\in }_0{h}^{2}n}$….. (iii)
Where $R=\frac{m{e}^4}{8{\in }_0^{2}c{h}^3}\mathrm{....}\left(iv\right)$
After substituting the values in (iii) & (iv), we get n= 25