Question

A hydrogen atom in ground state is moving with a kinetic energy of 30 eV. It collides with a deuterium atom in ground state at rest. The hydrogen atom is scattered at right angle to its original line of motion. Assume that energy of nth state in both the atoms is given by En=−13.6n2eV and the mass of deuterium is twice that of hydrogen. Write the maximum and minimum possible kinetic energy of deuterium after collision.

Answer 1

Explanation:

The potential energy of a hydrogen atom is zero in ground state. An electron is board to the nucleus with energy 13.6 eV.

Now we have to give energy of 13.6 eV. To cancel that energy.

Then additional 10.2 eV, is required to attain first excited state.

Total energy of an atom in the first excited state is $$= 13.6 eV + 10.2 eV = 23.8 eV$$