The wavelength of the K_$$\alpha$$ X-ray line is related to the atomic number of the element by Moseley's law, which can be expressed as:

$$ \sqrt{\nu} = A(Z - B) $$

Here, $$\nu$$ is the frequency of the emitted X-ray, $$Z$$ is the atomic number, and $$A$$ and $$B$$ are constants. Given that the wavelength $$\lambda$$ is inversely proportional to the frequency $$\nu$$ (since $$\nu = \frac{c}{\lambda}$$, where $$c$$ is the speed of light), we can rewrite this equation in terms of wavelength:

$$ \lambda \propto \frac{1}{(Z - B)^2} $$

Now, let's consider the ratio of the wavelengths for copper (Cu) and molybdenum (Mo):

$$ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{Z_{Mo} - B}{Z_{Cu} - B}\right)^2 $$

For the K_$$\alpha$$ X-ray line, the constant $$B$$ can be approximated as 1. Using the atomic numbers $$Z_{Cu} = 29$$ and $$Z_{Mo} = 42$$, we can plug these values into the ratio:

$$ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{42 - 1}{29 - 1}\right)^2 = \left(\frac{41}{28}\right)^2 $$

Calculating the above expression:

$$ \frac{41}{28} \approx 1.464 $$

$$ \left(1.464\right)^2 \approx 2.14 $$

Therefore, the ratio $$\frac{\lambda_{Cu}}{\lambda_{Mo}}$$ is close to 2.14. The correct option is:

Option B 2.14