The minimum kinetic energy required for ionization of a hydrogen atom is `E_(1)` in case electron is collided with hydrogen atom , it is `E_(2)` if the the hydrogen ion is collided and `E_(1)` when helium ion collided . Then.

Options

A. `E_(1)=E_(2)=E_(3)`
B. `E_(1) gt E_(2) gt E_(3)`
C. `E_(1) lt E_(2) lt E_(3)`
D. `E_(1) gt E_(3) gt E_(2)`

Answer

C

Explanation

According to the question, ionization occurs as a result of a completely inelastic collision, then we can write,
`" "mv_(0)-0=(m+m_(H))u`
Where m is the mass of incident particle, `m_(H)` the mass of hydrogen atom, `v_(0)` the initial velocity of incident particle, and u the final common velocity of the particle after collision. Prior to collision, the K.E. of the incident particle was,
`" "E_(0)=(mv_(0)^(2))/2`
The total kinetic energy after collision,
`E=((m+m_(H))u^(2)))/2=(m^(2)v_(0)^(2))/(2(m+m_(H)))`
The decrease in kinetic energy must be equal to ionization energy. Therefore,
`" "E_(1)=E_(0)-E=(m_(H)/(m+m_(H)))E_(0) i.e., E_(1)/E_(0)=1/((1+m/m_(H)))`
i.e., the greater the mass m, the smaller the fraction of initial kinetic energy that be used for ionization.