Question

If for Li$^{2+}$ ion, electron is in transition between energy levels such that sum of principal quantum numbers is 4 and difference is 2, then find the wavelength (in cm) emitted for transition between these energy levels.
[Given: $R = 1.1 \times 10^5\ \text{cm}^{-1}$]

• A. $114 \times 10^{-8}$ cm
• B. $1026 \times 10^{-8}$ cm
• C. $12.66 \times 10^{-8}$ cm
• D. $10^{-8}$ cm

Hint:

For hydrogen-like ions, wavelength is inversely proportional to $Z^2$.

Answer 1

The Correct Option is A

Step 1: Determine principal quantum numbers.
Let $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$
Solving:
\[ n_1 = 1,\quad n_2 = 3 \]
Step 2: Apply Rydberg equation for hydrogen-like ions.
For Li$^{2+}$, $Z = 3$
\[ \frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]
Step 3: Substitute values.
\[ \frac{1}{\lambda} = 1.1 \times 10^5 \times 9 \left(1 - \frac{1}{9}\right) \] \[ = 1.1 \times 10^5 \times 8 = 8.8 \times 10^5 \]
Step 4: Calculate wavelength.
\[ \lambda = \frac{1}{8.8 \times 10^5} = 1.14 \times 10^{-6}\ \text{cm} \]