Question

In a sample of $H$-atoms, electrons make a transition from ground state to a particular excited state where path length is five times of de-Broglie wavelength. Now excited electron makes the back transition to the ground state producing all the possible photons of different energies. If photon having 2${}^{nd}$ highest energy of this spectrum, is used to excite the electron of 2${}^{nd}$ excited state of $L{i}^{2+}$ ion, then the final excited state of $L{i}^{2+}$ ion is:

• A. 11${}^{th}$
• B. 12${}^{th}$
• C. 9${}^{th}$
• D. 10${}^{th}$

Answer 1

Answer: (B)

12${}^{th}$

Path length $=2\pi r$, where r is the radius of the orbit.

According to the question, $2\pi r=5\frac{h}{mv}$

$\Rightarrow mvr=5\frac{h}{2\pi }$

According to quantization of angular momentum

$\Rightarrow mvr=n\frac{h}{2\pi }$

On comparing above equations, we get $n=5$ i.e excited state

Second highest energy transtion takes place from $n=4$ to $n=1$

$E=-13.6(\frac{1}{1^2}-\frac{1}{4^2})$

For $L{i}^{2+}$, $E=-13.6\times 9(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Both of the above energies are equal, comparing above 2 energies,

We get $n_1=3$ and $n_2=12$.