A particle A of mass m and initial velocity v collides with a particle B of mass m2 which is at rest- The collision is head on- and elastic- The ratio of the de-Broglie wavelengths -A to -B after the collision is -

The correct option is C λ_Aλ_B=2 From conservation of linear momentum, mv= mv_B + m2 v_B v= v_A+ v_B2 Also v^2 = v_a^2 + v_B^22 (v_A+ ...

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Standard XII

Physics

Matter Waves

A particle ...

Question

A particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $λ_{A}$ to $λ_{B}$ after the collision is :

\frac{λ_{A}}{λ_{B}} = \frac{1}{2}

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\frac{λ_{A}}{λ_{B}} = \frac{1}{3}

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\frac{λ_{A}}{λ_{B}} = 2

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\frac{λ_{A}}{λ_{B}} = \frac{2}{3}

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Solution

The correct option is C $\frac{λ_{A}}{λ_{B}} = 2$
From conservation of linear momentum, $m v = m v_{B} + \frac{m}{2} v_{B}$

$v = v_{A} + \frac{v_{B}}{2}$

Also $v^{2} = v_{a}^{2} + \frac{v_{B}^{2}}{2}$

${(v_{A} + \frac{v_{B}}{2})}^{2} = v_{A}^{2} + {(\frac{v_{B}}{2})}^{2}$

$v_{A} v_{B} = \frac{v_{B}}{4}$

$v_{B} = 4 v_{A}$ and $m_{B} = \frac{m_{A}}{2}$

$P_{A} = m_{A} v_{A}$

$P_{B} = \frac{m_{A}}{2} \times 4 v_{A} = 2 P_{A}$

$\frac{λ_{A}}{λ_{B}} = \frac{P_{B}}{P_{A}} = \frac{2 P_{A}}{P_{A}} = 2$

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A particle A of mass m and initial velocity v collides with a particle B of mass m2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is :

The correct option is C λAλB=2From conservation of linear momentum, mv=mvB+m2vBv=vA+vB2Also v2=v2a+v2B2(vA+vB2)2=v2A+(vB2)2vAvB=vB4vB=4vA and mB=mA2PA=mAvAPB=mA2×4vA=2PAλAλB=PBPA=2PAPA=2

A particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $λ_{A}$ to $λ_{B}$ after the collision is :