The wavelength of spectral line obtained in the spectrum of $$Li^{2+}$$ ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

We need to find the wavelength of a spectral line in $$Li^{2+}$$ for a transition where the sum of levels is 4 and difference is 2.

First, we find the energy levels by solving $$n_1 + n_2 = 4$$ and $$n_2 - n_1 = 2$$, which gives $$n_2 = 3, n_1 = 1$$.

Next, we use the Rydberg formula for $$Li^{2+}$$ (Z = 3):

$$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R \times 9 \times \left(1 - \frac{1}{9}\right) = 9R \times \frac{8}{9} = 8R$$

Here, $$R = 1.097 \times 10^5 \text{ cm}^{-1}$$. Substituting this value gives $$\frac{1}{\lambda} = 8 \times 1.097 \times 10^5 = 8.776 \times 10^5 \text{ cm}^{-1}$$, and hence $$\lambda = \frac{1}{8.776 \times 10^5} = 1.14 \times 10^{-6} \text{ cm}$$.

Therefore, the wavelength is Option 3: $$1.14 \times 10^{-6}$$ cm.