An `alpha`-particle with a kinetic energy of `2.1eV` makes a head on collision with a hydrogen atom moving towards it with a kinetic energy of `8.4 eV`. The collision

Options

A. must be perfectly elastic
B. may be perfectly inelastic
C. may be inelastic
D. must be perfectly inelastic

Answer

C

Explanation

The maximum kinetic energy avaiable for transition to potential energy`//` exclination energy is:
`(1)/(2).(m_(alpha)m_(H))/(m_(alpha)+m_(H)).(V_(rel))^(2)`
`=(4m_(1),m_(H))/(5m).(v_(alpha)+v_(H))^(2) " "=(2m)/(5).(v_(alpha)^(2)+v_(H)^(2)+2v_(alpha)v_(H))`
`(2m)/(5)[(2.E_(alpha))/(4m)+(2E_(4))/(m)+2sqrt((2E.alpha)/(4m).(2E_(H))/(m))=(2)/(5)[(2.1)/(2)+2xx8.4+2xxsqrt(2.1xx8.4)]`
`=10.5 eV gt 10.2 eV`
Hence, inelastic collision is possible.