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.