A particle P collides inelastically with a singly ionized helium atom which is at rest. Initially helium atom (He+) is in ground state. After collision particle ‘P’ is found to be scattered at an angle $$90$$° with respect to it’s initial direction of motion, and before emitting any radiation, $$He+$$ and particle P have same kinetic energy. The $$He+$$ atom after collision was found to be in such an excited state that the maximum possible number of different wavelengths representing the successive transitions in between excited level to ground level is $$6$$. Mark the CORRECT $$option(s)$$.

Question

A particle P collides inelastically with a singly ionized helium atom which is at rest. Initially helium atom  (He+) is in ground state. After collision particle ‘P’ is found to be scattered at an angle  $$90$$° with respect to it’s initial direction of motion, and before emitting any radiation,  $$He+$$ and particle P have same kinetic energy. The  $$He+$$ atom after collision was found to be in such an excited state that the maximum possible number of different wavelengths representing the successive transitions in between excited level to ground level is $$6$$. Mark the CORRECT $$option(s)$$.

Infinity Learn · Accepted Answer

The initial and final situations are marked using Conservation Of Linear Momentum.In x direction, $$mu+0=0+Mvx$$⇒$$vx=muMIn$$ y direction, $$0+0+0=mv-Mvy$$⇒$$vy=mvMWe$$ are told,  $$KEm=KEM$$ in final state.⇒    $$12mv2=12M$$ $$V2$$ ​⇒    $$12mv2=12Mm2M2(u2+v2)$$ $$($$∵from fig, $$V=vx2+vy2=(muM)2+(mvM)2)$$​⇒ $$v2=mM(u2+v2)$$ ⇒    $$v2=mu2M$$−m , which is possible for  M>mHence particle P can be electron or neutronFrom n $$=$$ $$7$$ to n $$=$$ $$1$$, $$6$$ maximum possible wavelengths are there, of which n $$=$$ $$4$$ to n $$=$$ $$3$$ only belongs to visible spectrum.

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