JEE Physics 2025 Question with Solution

Given: Mass of the particle is $10^{-30} \, \text{kg}$, velocity is $2.21 \times 10^6 \, \text{m/s}$, and Planck's constant is $6.63 \times 10^{-34} \, \text{J.s}$.

Find: Which radiation the particle behaves closely like under matter wave consideration.

Use the de Broglie wavelength relation:

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

Substituting the given values:

$$\lambda = \frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^6}$$ $$\lambda = \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}}$$ $$\lambda \approx 3.00 \times 10^{-10} \, \text{m}$$

This is $0.3 \, \text{nm}$, which lies in the X-ray range. Therefore, under matter wave consideration, the particle behaves closely like X-rays.

The correct option is B.

Given: $m = 10^{-30} \, \text{kg}$, $v = 2.21 \times 10^6 \, \text{m/s}$, and $h = 6.63 \times 10^{-34} \, \text{J.s}$.

Find: The corresponding region of the electromagnetic spectrum.

First calculate the de Broglie wavelength:

$$\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^6}$$ $$\lambda = 3 \times 10^{-10} \, \text{m}$$

The X-ray region typically extends from about $10^{-11} \, \text{m}$ to $10^{-8} \, \text{m}$. Since $3 \times 10^{-10} \, \text{m}$ falls in this interval, the particle behaves like X-rays.

Hence, the correct option is B.