To find the matter wave behavior of the particle, we use the de Broglie wavelength formula:

$$\lambda = \frac{h}{mv}$$

where:

$$h = 6.63 \times 10^{-34} \, \mathrm{J \cdot s}$$ is Planck’s constant,

$$m = 10^{-30} \, \mathrm{kg}$$ is the mass of the particle, and

$$v = 2.21 \times 10^6 \, \mathrm{m/s}$$ is its velocity.

Follow these steps:

Substitute the values into the formula:

$$\lambda = \frac{6.63 \times 10^{-34}}{(10^{-30})(2.21 \times 10^6)}$$

Calculate the denominator:

$$10^{-30} \times 2.21 \times 10^6 = 2.21 \times 10^{-24}$$

Now compute the wavelength:

$$\lambda \approx \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}} \approx 3 \times 10^{-10} \, \mathrm{m}$$

This wavelength, roughly $$3 \times 10^{-10} \, \mathrm{m}$$ (or 0.3 nm), lies within the X-ray portion of the electromagnetic spectrum, since X-rays typically have wavelengths in the range of about 0.01 nm to 10 nm.

Therefore, under the matter wave consideration, the particle will behave closely like:

Option A: X-rays.