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Question

A particle A with a mass mA is moving with a velocity v and hits a particle B of mass mB at rest. If motion is one dimensional and take the collision is elastic, then the change in the de Broglie wavelength of the particle A is

A
h2mAv[(mA+mB)(mAmB)1]
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B
hmAv[(mAmB)(mA+mB)1]
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C
hmAv[(mA+mB)(mAmB)1]
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D
2hmAv[(mA+mB)(mAmB)+1]
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Solution

The correct option is C hmAv[(mA+mB)(mAmB)1]
According to law of conversation of momentum
mAv+mB×0=mAvA+mBvB
or mA(vvB)=mBvB ....(i)
According to law of conversation of klinetic energy
12mAv2=12mAv2A+12mBv2B

or mA(v2V2A)=mBv2B

or mA(vvA)(v+vA)=mBv2b ...(ii)

Dividing (ii) by (i), we get
v+vA=vBorv=vBvA ...(iii)

Solving (i) and (iii), we get
vA=(mAmBmA+mB)v and vB=(2mAmA+mB)v

λinitial=hmAv and λfinal=hmAvA=h(mA+mB)mA(mAmB)v

Δλ=λfinalλinitial=hmAv[(mA+mB)(mAmB)1]

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