QJEEM25Phc17f: JEE Main 2025 Physics - Class 12 JEEM25Phc17f Solution: A sub-atomic particle of mass is moving with a velocity . Under …

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.