Sol. Mass of neutron, m 1 1 u.

Mass of deuterium, m 2 2 u.

Using classical mechanics, we obtain

= = =

where E 0 is the initial kinetic energy of the neutron and Δ E is the energy loss.

After 1 st collision, Δ E 1 = E 0

After 2 nd collision, Δ E 2 = E 1

After 3 rd collision, Δ E 3 = E 2

After n th collision, Δ E n = E n – 1

Adding all the losses, we get

Δ E = Δ E 1 + Δ E 2 + … + Δ E n = (E 0 + E 1 + E 2 + … + E n –1 )

Since, E 1 = E 0 – Δ E 1 = E 0 and E 2 = E 1 = E 0 ,

E n – 1 = E 0

∴ Δ E = E 0

or = = 1 –

Here, E 0 = 1 × 10 6 eV

Δ E = 1 × 10 6 – 0.025 eV

∴ = = or 9 n = 4 × 10 7 Taking log of both sides and solving, we get n = 8.