A) \[\frac{hc}{\lambda }+\frac{1}{2}M{{u}^{2}}-\frac{1}{2}M{{v}^{2}}\]
B) \[\frac{1}{2}M{{u}^{2}}-\frac{1}{2}M{{v}^{2}}-\frac{hc}{\lambda }\]
C) \[\frac{1}{2}M{{u}^{2}}-\frac{hc}{\lambda }\]
D) \[\frac{hc}{\lambda }-\frac{1}{2}M{{u}^{2}}-\frac{1}{2}M{{v}^{2}}\]
Correct Answer: D
Solution :
[d] Momentum conservation v' = velocity of the second atom \[Mu=\frac{M}{2}v'\cos \theta \] \[\Rightarrow \,\,\,\,v'\cos \theta =2u\] ...(i) And \[\frac{M}{2}v'\sin \theta =\frac{M}{2}v\] \[\Rightarrow \,\,\,\,v'\sin \theta =v\] ...(ii) From (i) and (ii), \[v'=\sqrt{{{(2u)}^{2}}+{{v}^{2}}}\] Energy conservation \[\frac{1}{2}M{{u}^{2}}+\frac{hc}{\lambda }=BE+\frac{1}{2}\frac{M}{2}{{v}^{2}}+\frac{1}{2}\frac{M}{2}{{(v')}^{2}}\] \[\therefore \,\,\,\,BE=\frac{1}{2}M{{u}^{2}}+\frac{hc}{\lambda }-\frac{M}{4}{{v}^{2}}-\frac{M}{4}(4{{u}^{2}}+{{v}^{2}})\] \[=\frac{hc}{\lambda }-\frac{M{{u}^{2}}}{2}-\frac{M{{v}^{2}}}{2}\]You need to login to perform this action.
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