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Question:

Two particles move at right angles to each other. Their de-Broglie wavelengths are λ₁ and λ₂ respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength λ, of the final particle, is given by

λ=λ₁+λ₂/2

2λ=1/λ₁+1/λ₂

1/λ²=1/λ₁²+1/λ₂²

λ=λ₁λ₂

Solution:

Correct option is D. 1/λ²=1/λ₁²+1/λ₂²
P→₁=h/λ₁i^ and P→₂=h/λ₂j^
Using momentum conservation
P→=P→₁+P→₂=h/λ₁i^+h/λ₂j^
|P→|=(h/λ₁)2+(h/λ₂)2
h/λ=(h/λ₁)2+(h/λ₂)2
1/λ²=1/λ₁²+1/λ₂².