Particle A of mass $$mA=m2$$ moving along the $$x-axis$$ with velocity $$v0$$ collides elastically with another particle B at rest having mass $$mB=m3$$. If the particles move along the $$x-axis$$ after the collision, the change Δ λ in $$de-Broglie$$ wavelength of particle A, in terms of its $$de-Broglie$$ wavelength λ$$0$$ before collision is :

Question

Particle A of mass $$mA=m2$$  moving along the $$x-axis$$ with velocity  $$v0$$ collides elastically with another particle B at rest having mass  $$mB=m3$$. If the particles move along the $$x-axis$$ after the collision, the change Δ λ  in $$de-Broglie$$ wavelength of particle A, in terms of its $$de-Broglie$$ wavelength λ$$0$$  before collision is :

Infinity Learn · Accepted Answer

Conservation of momentum, ​$$m2V0+m30=m2VA+m3VB$$​⇒$$3VA+2VB=3V0$$....................$$.1$$​Coefficient of restitution​,$$e=1=VA$$−$$VB0$$−$$V0$$​⇒VA−$$VB=$$−$$V0$$......................$$.2Solving$$ $$1$$ and $$2$$​$$VA=V05$$​Now, de broglie wavelength $$=$$λ$$=hmvSo$$, Initial wavelength of $$A=$$λ$$i=hm2V0=$$λ$$0and$$ Final wavelength of $$A=$$λ$$f=hm2V05=5$$λ$$0Thus$$, Change in wavelength $$=$$△λ$$=$$λf−λ$$i=4$$λ$$0In$$ ground frame,  $$m2$$ moves forward with  $$3V05$$−$$2V05=V05So$$  λ$$f=5$$λ$$0$$⇒Δ λ$$=4$$λ$$0$$

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