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Q.

A particle is moving with a velocity of $10^{4} m / s .$ The product of uncertainity in position and velocity of this particle is equal to $\frac{10^{- 14}}{4 π} m^{2} / s .$ Debroglie wavelength of the particle would be

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a

$10^{- 22} m$

b

$10^{- 20} m$

c

$10^{- 16} m$

d

$10^{- 18} m$

answer is D.

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Detailed Solution

$∆ x . ∆ v = \frac{h}{4 πm}; Given, ∆ x . ∆ v = \frac{10^{- 14}}{4 π} m^{2} / s; \frac{10^{- 14}}{4 π} m^{2} / s = \frac{h}{4 πm}; 10^{- 14} m^{2} / s = \frac{h}{m}; . . . . . (1) we know λ = \frac{h}{mv}; . . . . . . (2) Given, v = 10^{4} m / s; Substituting (1) in (2) λ = \frac{10^{- 14} m^{2} / s}{10^{4} m / s} = 10^{- 18} m$

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