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Question
Two particles A and B of de Broglie wavelengths λ1 and λ2 combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).
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Solution
By the law of conservation of momentum,
|PC| = |PA| + |PB|
Let us first take the case I when both PA and PB are positive, PA
then `λ_C = (λ_Aλ_B)/(λ_A + λ_B)`
In the second case when both PA, and PB are negative,
then `λ_C = (λ_Aλ_B)/(λ_A + λ_B)`
In case III when PA > PB < 0 i.e., PA is positive and PB is negative,
`h/λ_C = h/λ_A - h/λ_B = ((λ_B - λ_A)h)/(λ_Aλ_B)`
⇒ λC = `(λ_Aλ_B)/(λ_B - λ_A)`
And In the case of IV when PA < 0, PB > 0, i.e., PA is negative and PB is positive.
∴ `h/λ_C = (-h)/λ_A + h/λ_B`
⇒ `((λ_A - λ_B)h)/(λ_Aλ_B)`
⇒ `λ_C = (λ_Aλ_B)/(λ_A - λ_B)`
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