Question

If Planck’s constant h = 6.6×10⁻34Js, the de Broglie wavelength of a particle having momentum 3.3×10⁻24kg m s⁻1 will be:

• A. \(0.002\,\text{\AA}\)
• B. \(0.5\,\text{\AA}\)
• C. \(2\,\text{\AA}\)
• D. 500AA

Hint:

Convert meters to angstrom carefully: 1AA=10⁻10m.

Answer 1

The Correct Option is C

Step 1: λ = (h)/(p)
Step 2: λ = frac6.6×10⁻343.3×10⁻24 = 2×10⁻10m = 2AA