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

If A is de-Broglie wavelength, then for nth stationary orbit

$2{\mathrm{\pi r}}_{\mathrm{n}}=\mathrm{n\lambda }$

where r_n is radius of nth orbit.

$\begin{array}{l}{\mathrm{r}}_{\mathrm{n}}=\frac{{\mathrm{\epsilon }}_0{\mathrm{h}}^2{\mathrm{n}}^2}{{\mathrm{\pi mZe}}^2}\\ \therefore 2\mathrm{\pi }\left(\frac{{\mathrm{\epsilon }}_0{\mathrm{h}}^2{\mathrm{n}}^2}{{\mathrm{\pi mZe}}^2}\right)=\mathrm{n\lambda }\Rightarrow \frac{1}{\mathrm{\lambda }}=\frac{{\mathrm{mZe}}^2}{2{\mathrm{\epsilon }}_0{\mathrm{h}}^2\mathrm{n}} ........\left(\mathrm{i}\right)\end{array}$

For Lyman series of hydrogen like atom, wavelength is given as,

$\frac{1}{\mathrm{\lambda }}={\mathrm{Z}}^2\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{{\mathrm{n}}^2}\right) ........\left(\mathrm{ii}\right)$

From Eqs. ) and (i), ${\mathrm{Z}}^2\mathrm{R}\left(1-\frac{1}{{\mathrm{n}}^2}\right)=\frac{{\mathrm{mZe}}^2}{2{\mathrm{\epsilon }}_0{\mathrm{h}}^2\mathrm{n}}$

Rydberg constant, $\mathrm{R}=\frac{{\mathrm{me}}^4}{8{\mathrm{\epsilon }}_0^2{\mathrm{ch}}^3}$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{{\mathrm{Z}}^2{\mathrm{me}}^4}{8{\mathrm{\epsilon }}_0^2{\mathrm{ch}}^3}\left(1-\frac{1}{{\mathrm{n}}^2}\right)=\frac{{\mathrm{mZe}}^2}{2{\mathrm{\epsilon }}_0{\mathrm{h}}^2\mathrm{n}}\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\left(1-\frac{1}{{\mathrm{n}}^2}\right)=\frac{4{\mathrm{\epsilon }}_0\mathrm{ch}}{{\mathrm{m}}^2\mathrm{Z}}\end{array}$

Substituting given values, we get

$\begin{array}{l}=\frac{4\times \left(8.85\times {10}^{-12}\right)\times \left(3\times {10}^8\right)\times \left(6.62\times {10}^{-34}\right)}{\mathrm{n}\times {\left(1.6\times {10}^{-19}\right)}^2\times 11}=\frac{25}{\mathrm{n}}\\ \therefore {\mathrm{n}}^2-1=25\mathrm{n}\\ \text{ or }{\mathrm{n}}^2-25\mathrm{n}-1=0\\ \mathrm{n}-\frac{25\pm \sqrt{(-25{)}^2+4\times 1\times 1}}{2}\approx \frac{25\pm \sqrt{625}}{2}\approx 25\end{array}$ 

As negative n is not possible, $\mathrm{n}\approx 25$

Alternative solution  :     Energy of emitted photon =

 = $$\begin{gathered} \frac{hC}{\lambda }= 13.6 z^2(1-\frac{1}{n^2}) \\ where \lambda = wavelength of emitted photon \\ =de broglie wavelength in n th orbit = \frac{3.3n}{z}\stackrel{0}{A} \end{gathered}$$

Solving n = 25