A hydrogen atom in ground state, moving with speed `v` collides with another hydrogen atom in ground state at rest. If `vlev_(0)=sqrt((kE_(0))/(m_(H)))~~a*bcxx10^(n)m//s`, then the collision is elastic. Here `a,b` and `c` are whole number, less then 9. Find the value of `(axxbxxc)/(kxxn)`
`(me^(4))/(2h^(2))=E_(0)=13.6eV//"atom"=2.18xx10^(-18)J//"atom" implies` ionisation energy of `H` -atom
`implies m_(H)=1.67xx10^(-27)kgimplies` Mass of hydrogen atom.

Answer

6

Explanation

According to Bohr's model
`DeltaE=E_(0)(1- 1/(n^(2)))impliesDeltaE_("min")=(3E_(0))/4`
During inelastic collision, a part kinetic energy of colliding particles is converted into internal energy. The internal energy of the system of two hydrogenatoms, considered in the problem cannot be changed less an `DeltaE_("min")`. It means if the change in kinetic eneryg of system in ground frame is less than `DeltaE_("min")`, then collision must be an elastic one. Hence cosidering the critical case
`1/2 m_(H)((v_(0))/2)^(2)xx2=(3E_(0))/2impliesv_(0)=sqrt((3E_(0))/(m_(H)))`
`sqrt((3xx2.18xx10^(-18))/(1.67xx10^(-27)))=sqrt(39.1617)xx10^(4)=6.257xx10^(4)m//s~~6.26xx10^(4)m//s`
`(axxbxxc)/(kxxn)=(6xx2xx6)/(3xx4)=6`