A deuteron of mass M moving at speed v collides elastically with an α -Particle of mass 2M, Initially at rest. The deuteron is scattered through 90° from initial direction of its motion with speed V d

Question

A deuteron of mass M moving at speed v collides elastically with an α -Particle of mass 2M, Initially at rest. The deuteron is scattered through 90° from initial direction of its motion with speed V d ​ while the α -particle is scattered with speed v α ​ at an angle θ from the initial direction of motion of deuteron. Then find ratio of D-broglie wave length of deuteron and α -Particle.

TestHub · Accepted Answer

By conservation of linear momentum:

(i) Along x-axis: $Mv + 0 = 2 MV_{α} cos θ \Rightarrow V = 2 V_{α} cos θ$

(ii) Along y-axis: $0 + 0 = 2 MV_{α} sin θ - MV_{d} \Rightarrow V_{d} = 2 V_{α} sin θ$

By conservation of kinetic energy: $\frac{1}{2} Mv^{2} = \frac{1}{2} MV_{d}^{2} + \frac{1}{2} (2 M) V_{α}^{2} \Rightarrow v^{2} = V_{d}^{2} + 2 V_{α}^{2}$

$(2 V_{α} cos θ)^{2} = (2 V_{α} sin θ)^{2} + 2 V_{α}^{2} \Rightarrow cos 2 θ = \frac{1}{2} \Rightarrow 2 θ = 6 0^{\circ}$ ⇒ $θ = 3 0^{\circ}, V_{d} = V_{α} = \frac{V}{3}$

De Broglie wavelengths: $λ_{d} = \frac{h}{MV _{d}}, λ_{α} = \frac{h}{( 2 M ) v _{α}}$ ⇒ $\frac{λ _{d}}{λ _{α}} = \frac{2 V _{α}}{V _{d}} = 2$ hence (D).

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