The de-Broglie wavelength of a particle of mass 6.63 g moving with a velocity of 100 ms$$^{-1}$$ is:

The de-Broglie wavelength $$\lambda$$ of a particle is given by the formula $$\lambda = \frac{h}{p}$$, where $$h$$ is Planck's constant and $$p$$ is the momentum of the particle. Momentum $$p$$ is calculated as $$p = m \times v$$, where $$m$$ is the mass and $$v$$ is the velocity. So, the formula becomes $$\lambda = \frac{h}{m v}$$.

We are given the mass $$m = 6.63$$ grams and velocity $$v = 100$$ meters per second. However, the SI unit for mass is kilograms, so we must convert grams to kilograms. Since 1 kilogram = 1000 grams, we have:

$$m = \frac{6.63}{1000} = 0.00663 \text{ kg}$$

This can also be written as $$m = 6.63 \times 10^{-3}$$ kg. The velocity $$v = 100$$ m/s remains as is. Planck's constant $$h$$ is $$6.63 \times 10^{-34}$$ Joule seconds (J s), and since Joule is kg m² s⁻², the units are consistent.

Now substitute the values into the formula:

$$\lambda = \frac{6.63 \times 10^{-34}}{m \times v} = \frac{6.63 \times 10^{-34}}{(6.63 \times 10^{-3}) \times 100}$$

First, compute the denominator: $$m \times v = (6.63 \times 10^{-3}) \times 100$$. Since $$100 = 10^2$$, we have:

$$m \times v = 6.63 \times 10^{-3} \times 10^2 = 6.63 \times 10^{-3 + 2} = 6.63 \times 10^{-1}$$

Now the expression is:

$$\lambda = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-1}}$$

Simplify by dividing the numerical coefficients and the powers of 10 separately. The numerical part is $$\frac{6.63}{6.63} = 1$$. The exponent part is $$\frac{10^{-34}}{10^{-1}} = 10^{-34 - (-1)} = 10^{-34 + 1} = 10^{-33}$$. Therefore:

$$\lambda = 1 \times 10^{-33} \text{ m}$$

So, the de-Broglie wavelength is $$10^{-33}$$ meters. Comparing with the options, this matches option A.

Hence, the correct answer is Option A.