The de-broglie wavelength of a particle as seen from an inertial frame is 25 μm and as seen from another inertial frame moving in opposite direction to the first (but with same speed) is 20 μm . The mass of particle is 6.63×10−30kg. What is its de-broglie wavelength (in μm )as seen from ground frame? The velocity of all three (particle and both frames) is along the same line. (Speed of particle is greater than either of frames)

λ1=hm(v1−v2)=25μ, λ2=hm(v1+v2)=20μm, ⇒v1+v2v1−v2=54⇒v2=v19again λ1=hm(v1−v19)=25μm⇒hmv1=25×89μm, λ=hmv1=2009μm=22.22μm

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The de-broglie wavelength of a particle as seen from an inertial frame is 25 $μ m$ and as seen from another inertial frame moving in opposite direction to the first (but with same speed) is 20 $μ m$ . The mass of particle is $6.63 \times 10^{- 30} k g .$ What is its de-broglie wavelength (in $μ m$ )as seen from ground frame? The velocity of all three (particle and both frames) is along the same line. (Speed of particle is greater than either of frames)

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answer is 22.22.

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Detailed Solution

$λ_{1} = \frac{h}{m (v_{1} - v_{2})} = 25 μ, λ_{2} = \frac{h}{m (v_{1} + v_{2})} = 20 μ m, \Rightarrow \frac{v_{1} + v_{2}}{v_{1} - v_{2}} = \frac{5}{4} \Rightarrow v_{2} = \frac{v_{1}}{9}$

again $λ_{1} = \frac{h}{m (v_{1} - \frac{v_{1}}{9})} = 25 μ m \Rightarrow \frac{h}{m v_{1}} = \frac{25 \times 8}{9} μ m, λ = \frac{h}{m v_{1}} = \frac{200}{9} μ m = 22.22 μ m$