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.

Options

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]`

Answer

C

Explanation

(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]`