A particle of mass M moving along the positive x-axis has a de Broglie wavelength ₀ . It decays into two identical fragments, each of mass M 2 . One of the fragments is observed to move along the positive y-axis with the same speed as the original particle. The de Broglie wavelength of the second fragment is:

Step-by-step solution

Let the initial velocity of the particle be v along the x-axis. Its initial momentum is p _i = Mv i . The initial de Broglie wavelength is ₀ = h Mv . After decay, one fragment of mass M 2 moves along the y-axis with speed v . Its momentum is p ₁ = M 2 v j . By conservation of linear momentum: p _i = p ₁ + p ₂ Mv i = M 2 v j + p ₂ p ₂ = Mv i - M 2 v j The magnitude of the momentum of the second fragment is: | p ₂| = (Mv)^2 + ( Mv 2 )^2 = Mv 1 + 1 4 = 5 2 Mv The de Broglie wavelength of the second fragment is: ₂ = h