question_answer

A diatomic molecule moving-at a speed u absorbs a photon of wavelength \[\lambda \] and then dissociates into two identical atoms. One of the atoms is found to be moving with a speed v in a direction perpendicular to the initial direction of motion of the molecule. Take mass of the molecule to be M and find the binding energy of the molecule. Assume that momentum of absorbed photon is negligible compared to that of the molecule.

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 \[\frac{1}{2}M{{u}^{2}}+\frac{hc}{\lambda }=BE+\frac{1}{2}\frac{M}{2}{{v}^{2}}+\frac{1}{2}\frac{M}{2}{{(v')}^{2}}\] \[\therefore \,\,\,\,BE=\frac{1}{2}M{{u}^{2}}+\frac{hc}{\lambda }-\frac{M}{4}{{v}^{2}}-\frac{M}{4}(4{{u}^{2}}+{{v}^{2}})\] \[=\frac{hc}{\lambda }-\frac{M{{u}^{2}}}{2}-\frac{M{{v}^{2}}}{2}\]