Answer to Question #98726 in Classical Mechanics for mariam

Question

An H2 molecule can be formed in a collision that involves three hydrogen atoms. Suppose that before such a collision, each of the three atoms has speed 4000 m/s , and they are approaching at 120 ∘ angles so that at any instant, the atoms lie at the corners of an equilateral triangle. Find the speeds of the H2 molecule and of the single hydrogen atom that remains after the collision. The binding energy of H2 is Δ=7.23×10−19J, and the mass of the hydrogen atom is 1.67×10−27kg.

Answer 1

The total momentum of the atoms before the collision is 0, therefore, after the collision, the atom and molecule will move in opposite directions with speeds that differ by half. From the law of conservation of energy it follows that the speed of an atom

$u^2 = 2v^2 + \frac{4\Delta}{3m}$ , or $u = (10^8(0.32 + 5.8))^{0.5} = 25 (km/s),$ and the speed of the molecule is 12.5 km/s.