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:
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:
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Convert meters to angstrom carefully: 1AA=10⁻10m.
Updated On: Mar 20, 2026
- \(0.002\,\text{\AA}\)
- \(0.5\,\text{\AA}\)
- \(2\,\text{\AA}\)
- 500AA
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The Correct Option is C
Solution and Explanation
Step 1: λ = (h)/(p)
Step 2: λ = frac6.6×10⁻343.3×10⁻24 = 2×10⁻10m = 2AA
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