Sol. Let u be the velocity of the hydrogen atom before collision and v the velocity of the two atoms moving together after collision. By the particle of conservation of momentum, we have:

Mu + M × 0 = 2Mv or v =

The loss in kinetic energy Δ E due to collision is given by

Δ E = Mu 2 – (2M)v 2 As v = u/2, we have Δ E = Mu

2 – (2M)(u/2) 2

= Mu 2 – Mu 2 = Mu 2 This loss in energy is due to the excitation of one of the hydrogen atoms. The ground state (n = 1) energy of a hydrogen atom is:

E 1 = – 13.6 eV

The energy of the first excited level (n = 2) is

E 2 = – 3.4 eV

Thus, the minimum energy required to excite a

hydrogen atom from ground state of first excited state is E 2 – E 1 = [–3.4 – (–13.6)]eV = 10.2 eV

= 10.2 × 1.6 × 10

–19 J

= 16.32 × 10 –19 J

As per the problem, the loss in kinetic energy in

collision is due to the energy used up in exciting one of

the atoms. Thus

Δ E = E 2 – E 1 or Mu 2 = 16.32 × 10 –19

or u 2 =

The mass of the hydrogen atom is 1.0078 amu or

1.0078 × 1.66 × 10 –27 kg.

∴ u 2 = = 39.02 × 10 8 or u = 6.246 × 10

4 ms –1