An $$\alpha$$ particle and a carbon 12 atom has same kinetic energy $$K$$. The ratio of their de-Broglie wavelengths $$(\lambda_\alpha : \lambda_{C12})$$ is

We need to find the ratio of de-Broglie wavelengths of an alpha particle and a Carbon-12 atom, both having the same kinetic energy $$K$$.

The de-Broglie wavelength is given by:

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$

For the alpha particle (mass number 4):

$$\lambda_\alpha = \frac{h}{\sqrt{2 \cdot 4u \cdot K}}$$

For the Carbon-12 atom (mass number 12):

$$\lambda_{C12} = \frac{h}{\sqrt{2 \cdot 12u \cdot K}}$$

Taking the ratio:

$$\frac{\lambda_\alpha}{\lambda_{C12}} = \frac{\sqrt{2 \cdot 12u \cdot K}}{\sqrt{2 \cdot 4u \cdot K}} = \sqrt{\frac{12}{4}} = \sqrt{3}$$

Therefore, $$\lambda_\alpha : \lambda_{C12} = \sqrt{3} : 1$$.

The correct answer is Option A.