The wavelength of de-Broglie wave is 2 μ m then its momentum is ( h = 6.63 × 10 - 34 J s )

Question

The wavelength of de-Broglie wave is 2 μ m then its momentum is ( h = 6.63 × 10 - 34 J s ), 3.315 × 10 - 28 k g m s - 1, 1.66 × 10 - 28 k g m s - 1, 4.97 × 10 - 28 k g m s - 1, 9.9 × 10 - 28 k g m s - 1

MARKS App · Accepted Answer

According to de-Broglie hypothesis λ = h p Where, h is Planck's constant. ⇒ p = h λ Given, h = 6.63 × 10 - 34 J-s λ = 2 μ m = 2 × 10 - 6 m ∴ p = 6.63 × 10 - 34 2 × 10 - 6 = 3.315 × 10 - 28 kg m s - 1

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