Question

An electron makes transition from higher energy orbit to lower energy orbit in $Li^{2+}$ ion such that $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$. Determine the wavelength of emitted photon in transition (in cm).

• A. $1.14 \times 10^{-6}$ cm
• B. $3.28 \times 10^{-6}$ cm
• C. $5.76 \times 10^{-6}$ cm
• D. $8.23 \times 10^{-6}$ cm

Hint:

For hydrogen-like ions, energy levels depend on $Z^2$. Always include nuclear charge.

Answer 1

The Correct Option is A

Step 1: Finding quantum numbers.
Given:
\[ n_1 + n_2 = 4, \quad n_2 - n_1 = 2 \]
Solving,
\[ n_2 = 3, \quad n_1 = 1 \]
Step 2: Energy difference for hydrogen-like ion.
For $Li^{2+}$, $Z = 3$.
\[ \Delta E = 13.6 Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right) \]
\[ \Delta E = 13.6 \times 9 \left(1 - \dfrac{1}{9}\right) = 108.8 \text{ eV} \]
Step 3: Calculating wavelength.
\[ \lambda = \dfrac{hc}{\Delta E} \]
Using $hc = 12400$ eV\AA :
\[ \lambda = \dfrac{12400}{108.8} \,\text{\AA} = 114 \,\text{\AA} \]
\[ \lambda = 1.14 \times 10^{-6} \text{ cm} \]