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

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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. $\Rightarrow$ 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 = $\infty$ 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 $\times$ (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.$\frac{6}{2}$^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

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