Question

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)

• A. Infra-red radiation
• B. X-rays
• C. Gamma rays
• D. Visible radiation

Hint:

The de Broglie wavelength can be used to calculate the behavior of particles as waves. If the wavelength is on the order of \( 10^{-10} \) m, the particle behaves like X-rays.

Answer 1

The Correct Option is B

To determine how a sub-atomic particle behaves under matter wave consideration, we can use the concept of de Broglie wavelength. The de Broglie wavelength \(\lambda\) of a particle is given by the formula:

\(\lambda = \frac{h}{mv}\) 

where:

• \(h = 6.63 \times 10^{-34}\) is the Planck's constant (in Joule-seconds).
• \(m = 10^{-30}\) kg is the mass of the particle.
• \(v = 2.21 \times 10^6\) m/s is the velocity of the particle.

Substituting these values into the formula, we get:

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

Calculating the value:

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

\(\lambda \approx 3.00 \times 10^{-10}\) meters or 0.3 nanometers.

This wavelength value is in the vicinity of X-rays, which typically range from about 0.01 to 10 nanometers. Thus, under the matter wave consideration, the particle behaves closely like X-rays.

Conclusion: The particle behaves like X-rays under matter wave consideration.

Answer 2

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:

• \( \lambda \) is the de Broglie wavelength
• \( h = 6.63 \times 10^{-34} \) J.s is Planck's constant
• \( m = 10^{-30} \) kg is the mass of the particle
• \( v = 2.21 \times 10^6 \) m/s is the velocity of the particle

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.