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)/(2m_(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. `(2h)/(m_(A^(v)))[((m_(A)+m_(B)))/((m_(A)-m_(B)))+1]` |
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Answer» Correct Answer - C (c ) : According to law of conservation of momentum `m_(A)^(v)+m_(B)xx0m_(A)v_(A)+m_(B)v_(B)` or `m_(A)(v-v_(A))=m_(B)v_(B)` . . .(i) According to law of conservation of kinetic energy `(1)/(2)m_(A)v^(2)=(1)/(2)m_(A)^(2)+(1)/(2)m_(B)v_(B)^(2)` or `m_(A)(v^(2)-v_(A)^(2))=m_(B)v_(B)^(2)` or `m_(A)(v-v_(A))(v+v_(A))=m_(B)v_(B)^(2)` or `m_(A)(v-v_(A))(v+v_(A))=m_(B)v_(B)^(2)` . . . (ii) Dividing (ii) by (i), we get `v+v_(A)=v_(B) or v=v_(B)-v_(A)` . . . (iii) Solving (i) and (iii), we get `v_(A)=((m_(A)-m_(B))/(m_(A)+m_(B)))v and v_(B)=((2m_(A))/(m_(A)+m_(B)))v` `lamda_("initial")=(h)/(m_(A)v) and lamda_("final")=(h)/(m_(A)v_(A))=(h(m_(A)+m_(B)))/(m_(A)(m_(A)-m_(B))v)` `:.Deltalamda=lamda_("final")-lamda_("initial")=(h)/(m_(A)v)[((m_(A)+m_(B)))/((m_(A)+m_(B)))-1]` |
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