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 can be used to excite the electron in a particular excited state of $L{i}^{2+}$ atom then find the final excited state of $L{i}^{2+}$ atom

Answer 1

From de- broglie wavelength and bohr's theory, we know
$2\pi r=n\lambda$.
In the particular excited state of H-atom, path length is five times the de-broglie wavelength.
$\therefore 2\pi r=5\lambda$ ...(1)

However, path length in a state n is n times the de-Broglie wavelength.
$\therefore 2\pi r=n\lambda$ .....(2)

From (1) and (2), principal quantum number (n) of the excited state = 5. Photon having 2nd highest energy corresponds to back transition of electron from n = 4 to n = 1
This photon will cause an already excited $L{i}^{2+}$ electron to go to some higher state. Let, the initial excited state of $L{i}^{2+}$ ion be $n_1$ and final excited state of $L{i}^{2+}$ ion be $n_2$

$\therefore$$\underset{Photonhaving2ndhighestenergycorrespondingtotransitionn=4ton=1inH-atom}{\underset{}{13.6(1{)}^2[\frac{1}{1^2}-\frac{1}{4^2}]}}=\underset{EnergyabsorbedbyL{i}^{2+}iontomakeatransitionfrom{n}_{1}to{n}_2}{\underset{}{13.6(3{)}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]}}$

$\therefore 13.6[\frac{1}{1^2}-\frac{1}{4^2}]=13.6[\frac{3^2}{n_1^2}-\frac{3^2}{n_2^2}]or13.6[\frac{1}{1^2}-\frac{1}{4^2}]=13.6[\frac{1}{(n_1/3{)}^2}-\frac{1}{(n_2/3{)}^2}]$

On comparing both sides,
$\frac{n_1}{3}=1$ & $\frac{n_2}{3}=4$
so, $n_1=3$ and $n_2=12$
Thus, the final excited state of $L{i}^{2+}$ ion is n = 12.