Question

In a sample of hydrogen atom in ground state electrons make transition from ground state to a particular excited state where path length is five times de-broglie wavelength, electrons make back transition to the ground state producing all possible photons. If photon having 2nd highest energy of this sample used to excite the electron in a particular excited state of $L{i}^{2+}$ ion then find the final excited state of $L{i}^{2+}$ ion

Answer 1

Since electron goes the state where the path length is 5 times de-broglie wavelength.
$⟹2\pi r=5\lambda$
also $\frac{2\pi r}{n}=\lambda ⟹n=5$
Hence electron goes to the 5th state.
2nd highest energy line will be $4\to 1$
If this photon is used for $L{i}^{2+}$ then
$13.6(1-\frac{1}{4^2})=13.6\times z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$⟹\stackrel{n_1}{3}\to \stackrel{n_2}{12}$