InterviewSolution
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InterviewSolution
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A particle A with a mass `m_(A)` is moving with a velocity v and hits a particle B (mass `m_(B)`) at rest (one dimensional motion). Find the change in the de-Broglie wavelength of the particle A. Treat the collision as elastic.A. `h/(m_(A)v)[(m_(A)+(m_(B)))/(m_(A)-(m_(B)))-1]`B. `h/(m_(A)v)[(m_(A)-(m_(B)))/(m_(A)+(m_(B)))-1]`C. `h/(m_(A)v)[(m_(A)(m_(B)))/(m_(A)+(m_(B)))-1]`D. `h/(m_(A)v)[(m_(A)(m_(B)))/(m_(A)-(m_(B)))-1]` |
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Answer» Correct Answer - A From the law of conservation of momentum,`m_(A)v=m_(A)v_(A)+m_(B)v_(B)` or `m_(A)(v-v_(A))=m_(B)v_(B)`...(i) (as particle `B` is at rest, its initial velocity is zero and `v_(A)` and `v_(B)` are the velocities of particles `A` and `B` after collision) Since the collision is elastic, kinetic energy is conserved during collision `therefore1/2m_(A)v^(2)=1/2m_(A)v_(A)^(2)+1/2m_(B)v_(B)^(2)` `m_(A)(v^(2)-v_(A)^(2))=m_(B)v_(B)^(2)`..(ii) Dividing eqn.(ii) by eqn.(i), we obtain `(m_(A)(v^(2)-v_(A)^(2))=m_(B)v_(B)^(2))/(m_(A)(v-v_(A))=m_(B)v_(B)` or `v+v_(A)=v_(B)`...(iii) From eqns.(i) and (iii), `m_(A)(v-v_(A))=m_(B)(v+v_(A))` or `(m_(A)-m_(B))v=(m_(A)-m_(B))v_(A)` or `v/v_(A)=((m_(A)+m_(B))/(m_(A)-m_(B)))`..(iv) Initial wavelength of the particle `B`,i.e., `(lambda_(A))_(f)=h/(m_(A)v_(A))` Change in wavelength, `Delta lambda=(lambda_(A))_(f)-(lambda_(A))_(i)=h/(m_(A)v_(A))-h/(m_(A)v)` or `Delta lambda=h/(m_(A)v)[v/v_(A)-1]`..(v) From eqns. (iv) and (v), `Delta lambda=h/(m_(A)v)[((m_(A)+m_(B))/(m_(A)-m_(B)))-1]` |
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