An electron (m = 9.1 ｘ 10⁻³¹ kg) traveling at 2 ｘ 10⁷ m s⁻¹ undergoes a head on collision with a hydrogen atom (m = 1.67 ｘ 10⁻²⁷kg), which is initially at rest. Assume the collision to be perfectly elastic and motion to be a straight line, find the velocity of hydrogen atom. 

(Ans: 2.2 ｘ 10⁴ m s⁻¹)



Given:

Mass of electron = m₁ = 9.1 ｘ 10⁻³¹ kg

Initial velocity of electron = v₁ = 2 ｘ 10⁷ m s⁻¹

Mass of Hydrogen = m₂ = 1.67 ｘ 10⁻²⁷kg

Initial velocity of Hydrogen = v₂ = 0 m s⁻¹ (at rest)


To Find:

Final velocity of Hydrogen = v₂' =  ?


Solution:


The formula to find Final velocity  v₂' of Hydrogen of mass m₂ is:

v₂' = `\frac {(m₂-m₁)v₂}{m₁+m₂}` + `\frac {2m₁v₁}{m₁+m₂}`

by putting values

v₂' = `\frac {(1.67 ｘ 10⁻²⁷kg - 9.1 ｘ 10⁻³¹ kg) (0 m s⁻¹)}{9.1 ｘ 10⁻³¹ kg + 1.67 ｘ 10⁻²⁷kg}` + `\frac {2 ï½˜ 9.1 ｘ 10⁻³¹ kg x (2 ｘ 10⁷ m s⁻¹) }{9.1 ｘ 10⁻³¹ kg+ 1.67 ｘ 10⁻²⁷kg}`

v₂' = 0 + `\frac {2 ï½˜ 9.1 ｘ 10⁻³¹ kg x (2 ｘ 10⁷ m s⁻¹) }{9.1 ｘ 10⁻³¹ kg + 16700 ｘ 10⁻³¹kg}`

v₂' = `\frac {36.4ｘ 10⁻²⁴ kg m s⁻¹}{16709.1 ｘ 10⁻³¹ kg }`

v₂' = 0.002178 ï½˜ 10⁷  m s⁻¹

or

v₂' = 2.178 ï½˜ 10⁴ m s⁻¹ -------------Ans.


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