99 Percentile Qs Bank for JEE MainPhysicsAtomic Physics
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 ionised. The mass of a hydrogen atom = 1.67 × 10 - 27 kg . Assume completely inelastic collision.
Options
- A7 .2 × 10 4 m s - 1
- B2 .3 × 10 4 m s - 1
- C2 .3 × 10 5 m s - 1
- D7 .2 × 10 5 m s - 1
Correct answer
A. 7 .2 × 10 4 m s - 1
Step-by-step solution
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. ∆ E = 13.6 eV Assuming completely inelastic collision. For an inelastic collision v 1 = v 2 = v By momentum conservation, m v = 2 m V ⇒ V = v 2 1 2 m v 2 = 1 2 ⋅ 2 m V 2 + ∆ E = m v 2 2 + ∆ E ∵ V = v / 2 1 4 m v 2 = ∆ E v = 4 ∆ E m v m i n = 4 × 13.6 × 1.6 × 10 - 19 1.67 × 10 - 27 = 7.2 × 10 4 m/s