A sub-atomic particle of mass $$10^{-30}$$kg is moving with a velocity $$2.21\times10^{6}$$ m/s . Under the matter wave consideration, the particle will behave closely like
$$\left(h=6.63\times10^{-34}J.s\right)$$

We need to find the de Broglie wavelength of the particle and identify which type of electromagnetic radiation it corresponds to.

The de Broglie wavelength formula is $$\lambda = \frac{h}{mv}$$.

Since the mass of the particle is $$m = 10^{-30}$$ kg, its speed is $$v = 2.21 \times 10^{6}$$ m/s, and Planck’s constant is $$h = 6.63 \times 10^{-34}$$ J·s, substituting these values yields:

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

$$\lambda = \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}}$$

$$\lambda = 3 \times 10^{-10} \text{ m} = 3 \text{ Å}$$

Now, to classify this wavelength, note that visible radiation spans $$4000 - 7000$$ Å; infrared radiation corresponds to greater than $$7000$$ Å; X-rays are in the range $$0.1 - 100$$ Å; and gamma rays are below $$0.1$$ Å.

Since $$\lambda = 3$$ Å falls in the X-ray range, the particle will behave closely like X-rays. Therefore, the correct answer is Option 4: X-rays.