Question

A subatomic particle of mass \(2.2 \times 10^{-28}\) kg. Velocity = \(3 \times 10^5 \, \text{ms}^{-1}\). De Broglie wavelength?

• A. \( 4.0 \times 10^{-10} \, \text{m} \)
• B. \( 5.0 \times 10^{-10} \, \text{m} \)
• C. \( 6.0 \times 10^{-10} \, \text{m} \)
• D. \( 7.0 \times 10^{-10} \, \text{m} \)

Hint:

To calculate the De Broglie wavelength of a particle, use the formula \(\lambda = \frac{h}{mv}\), which relates the wavelength to the particle’s mass and velocity.

Answer 1

The Correct Option is B

The De Broglie wavelength \(\lambda\) of a particle is given by: \[ \lambda = \frac{h}{mv} \] Where: - \(h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\) is Planck’s constant, - \(m = 2.2 \times 10^{-28} \, \text{kg}\) is the mass of the particle, - \(v = 3 \times 10^5 \, \text{ms}^{-1}\) is the velocity of the particle. Substitute the given values into the formula: \[ \lambda = \frac{6.626 \times 10^{-34}}{(2.2 \times 10^{-28})(3 \times 10^5)} = 5.0 \times 10^{-10} \, \text{m} \] Thus, the De Broglie wavelength is \(5.0 \times 10^{-10} \, \text{m}\).