Question

The de Broglie wavelength of a particle of mass 1 mg moving with a velocity of \( 10 \) ms\(^{-1}\) is (h = \( 6.63 \times 10^{-34} \) Js)

• A. \( 6.63 \times 10^{-29} \) m
• B. \( 6.63 \times 10^{-31} \) m
• C. \( 6.63 \times 10^{-34} \) m
• D. \( 6.63 \times 10^{-22} \) m

Hint:

The de Broglie wavelength is inversely proportional to mass and velocity. Higher mass or velocity leads to a smaller wavelength.

Answer 1

The Correct Option is A

The de Broglie wavelength is given by: \[ \lambda = \frac{h}{m v} \] where: - \( h = 6.63 \times 10^{-34} \) Js (Planck’s constant) - \( m = 1 \) mg = \( 1 \times 10^{-6} \) kg - \( v = 10 \) m/s Substituting the values: \[ \lambda = \frac{6.63 \times 10^{-34}}{(1 \times 10^{-6}) \times (10)} \] \[ = \frac{6.63 \times 10^{-34}}{10^{-5}} \] \[ = 6.63 \times 10^{-29} \text{ m} \] Thus, the correct answer is \( 6.63 \times 10^{-29} \) m.