A sub-atomic particle of mass \( 10^{-30} \) kg is moving with a velocity of \( 2.21 \times 10^6 \) m/s. Under the matter wave consideration, the particle will behave closely like (h = \( 6.63 \times 10^{-34} \) J.s)
To determine the type of radiation with which the sub-atomic particle behaves closely, we need to calculate its de Broglie wavelength. This is given by the equation:
\(\lambda = \frac{h}{mv}\)
where:
Substituting the values into the de Broglie wavelength equation:
\(\lambda = \frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^6}\)
Solving for \( \lambda \):
\(\lambda = \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}}\)
\(\lambda = 3 \times 10^{-10} \) m
This wavelength corresponds to the X-ray region of the electromagnetic spectrum, which typically ranges from \( 10^{-11} \) m to \( 10^{-8} \) m. As a result, the sub-atomic particle behaves like X-rays.
Thus, the correct answer is: X-rays.