Two hydrogen atom in ground state are moving on opposite direction with the same speed and collide head on . The minimum kinetic energy of each hydrogen atom for the collision to be inelastic so that both the atoms are eccited isA. `13.6eV`B. `10.2eV`C. `20.4eV`D. `27.2eV`

Two hydrogen atom in ground state are moving on opposite direction wit

1.	Two hydrogen atom in ground state are moving on opposite direction with the same speed and collide head on . The minimum kinetic energy of each hydrogen atom for the collision to be inelastic so that both the atoms are eccited isA. `13.6eV`B. `10.2eV`C. `20.4eV`D. `27.2eV`
Answer» Correct Answer - B Let K by the initial energy of each atoms . Maximum loss in Ke takes place in perfectly inelastic collision. From conservation of linear momentum `"mu"+(-"mu")=2mvRightarrowv=0` Where v=common velocity after collision. therefore maximum possible loss in KE is ` DeltaK=1/2"mu"^(2)xx2-1/22mxx0^(0)=21/2"mu"^(2)=2K` Minimum energy required to excite both the electrons is `DeltaE=10.2+10.2=20.4eV`. For inelastic collision with two electron excited `DeltageDeltaERightarrow2Kge20.4RightarrowK_(min)=10.2eV`

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Two hydrogen atom in ground state are moving on opposite direction with the same speed and collide head on . The minimum kinetic energy of each hydrogen atom for the collision to be inelastic so that both the atoms are eccited isA. `13.6eV`B. `10.2eV`C. `20.4eV`D. `27.2eV`

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