= \frac{h}{p} $$

where $ h $ is Planck's constant and $ p $ is the momentum of the particle.

The momentum $ p $ can be expressed as:

$$ p = \sqrt{2mK} $$

where $ m $ is the mass and $ K $ is the kinetic energy of the particle.

Let's denote the masses and kinetic energies of particles $$\mathrm{A}$$ and $$\mathrm{B}$$ as follows:

Mass of particle $$\mathrm{A}$$: $$ m_A $$

Mass of particle $$\mathrm{B}$$: $$ m_B $$

Since the mass of particle $$\mathrm{A}$$ is double that of particle $$\mathrm{B}$$:

$$ m_A = 2m_B $$

Kinetic energy of particle $$\mathrm{A}$$: $$ K_A $$

Kinetic energy of particle $$\mathrm{B}$$: $$ K_B $$

Given that the kinetic energy of particle $$\mathrm{B}$$ is $$ \frac{1}{8} $$th that of $$\mathrm{A}$$:

$$ K_B = \frac{K_A}{8} $$

Now, let's find the ratio of the de Broglie wavelengths of particles $$\mathrm{A}$$ and $$\mathrm{B}$$. The de Broglie wavelength for particle $$\mathrm{A}$$ is:

$$ \lambda_A = \frac{h}{\sqrt{2m_A K_A}} $$

The de Broglie wavelength for particle $$\mathrm{B}$$ is:

$$ \lambda_B = \frac{h}{\sqrt{2m_B K_B}} $$

We need to find the ratio $$ \frac{\lambda_A}{\lambda_B} $$:

$$ \frac{\lambda_A}{\lambda_B} = \frac{\frac{h}{\sqrt{2m_A K_A}}}{\frac{h}{\sqrt{2m_B K_B}}} = \frac{\sqrt{2m_B K_B}}{\sqrt{2m_A K_A}} = \sqrt{\frac{m_B K_B}{m_A K_A}} $$

Substitute the given values:

$$ \frac{\lambda_A}{\lambda_B} = \sqrt{\frac{m_B \cdot \frac{K_A}{8}}{2m_B \cdot K_A}} = \sqrt{\frac{\frac{m_B K_A}{8}}{2m_B K_A}} = \sqrt{\frac{1}{16}} = \frac{1}{4} $$

Thus, the ratio of the de Broglie wavelength of $$ \mathrm{A} $$ to that of $$ \mathrm{B} $$ is:

Option C: 1 : 4

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