Question

A hydrogen atom moving at speed `v` collides with another hydrogen atom kept at rest .Find the minimum value of `v` for which one of the atoms may get ionized, the mass of a hydrogen atom `= 1.67 xx 10^(-27)kg`

Answer 1

Inelastic collision will take place if a part of incident kinetic energy is utilised in exciting the atom. Here one atom is to be ionised i.e. `DeltaE=13.6 eV`.
Assuming completely inelastic collision.
By momentum conservation
`m upsilon=2 mV implies V=v//2`
`(1)/(2)mv^(2)=(1)/(2).2 mV^(2)+DeltaE`
`m(v//2)^(2)+DeltaE`
`(1)/(4)mv^(2)=DeltaE`
`v=sqrt(4DeltaE)/(m)`
`v_(min)=sqrt((4xx13.6xx1.6xx10^(-19))/(1.67xx10^(-27)))`
`=7.2xx10^(8) m//sec`