Question

Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is ______.

Options

• 10.20 eV

• 20.40 eV

• 13.6 eV

• 27.2 eV

Answer 1

Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is 10.20 eV.

Explanation:

Total energy (E) is the sum of potential energy and kinetic energy, i.e. E = K + U.

⇒ `E = (kZe^2)/(2r_n)` also `r_n = (n^2h^2ε_0)/(pimze^2)`

Hence `E = - ((me^4)/(8ε_0^2h^2)) * Z^2/n^2 = - ((mw^4)/(8ε_0^2ch^3)) ch Z^2/n^2`

= `- R  ch Z^2/z^2`

= `- 13.6 Z^2/n^2 eV`

The lowest state of the atom, called the ground state, is that of the lowest energy. The energy of this state (n – 1), E, is – 13.6 eV.

Energy level diagram of hydrogen/hydrogen like atom:

Principal quantum number	Orbit	Excited state	Energy for H2-atom
n = ∞	Infinite	Infinite	0 eV
n = 4	Fourth	Third	– 0.85 eV
n = 3	Third	Second	– 1.51 eV
n = 2	Second	First	– 3.4 eV
n = 1	First	Ground	– 13.6 eV

Let two H atoms initially at in the ground state. Now two atoms collide inelastically. The total energy associated with the two H atoms = 2 × (13.6 eV) = 27.2 eV

The maximum amount by which their combined kinetic energy is reduced when  any one of them goes into the first excited state (n = 2) after the inelastic collision.

‘The total energy associated with the two H-atoms after the collision = `(13.6/2^2) + (13.6)` = 17.0 eV

‘The total energy associated with the two H-atoms after the collision = 27.2 – 17.0 = 10.2 eV