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

<p>To find the matter wave behavior of the particle, we use the de Broglie wavelength formula:</p> <p>$\lambda = \frac{h}{mv}$</p> <p>where:</p> <p><p>$h = 6.63 \times 10^{-34} \, \mathrm{J \cdot s}$ is Planck’s constant,</p></p> <p><p>$m = 10^{-30} \, \mathrm{kg}$ is the mass of the particle, and</p></p> <p><p>$v = 2.21 \times 10^6 \, \mathrm{m/s}$ is its velocity.</p></p> <p>Follow these steps:</p> <p><p>Substitute the values into the formula:</p> <p>$\lambda = \frac{6.63 \times 10^{-34}}{(10^{-30})(2.21 \times 10^6)}$</p></p> <p><p>Calculate the denominator:</p> <p>$10^{-30} \times 2.21 \times 10^6 = 2.21 \times 10^{-24}$</p></p> <p><p>Now compute the wavelength:</p> <p>$$\lambda \approx \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}} \approx 3 \times 10^{-10} \, \mathrm{m}$$</p></p> <p>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.</p> <p>Therefore, under the matter wave consideration, the particle will behave closely like:</p> <p>Option A: X-rays.</p>