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 n^(th) state in both the atoms is given by E_(n) = -(13.6)/(n^(2)) eV and the mass of deuterium is twice that of hydrogen. Write the maximum and minimum possible kinetic energy of deuterium after collision.

A hydrogen atom in ground state is moving with a kinetic energy of 30

1.	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 n^(th) state in both the atoms is given by E_(n) = -(13.6)/(n^(2)) eV and the mass of deuterium is twice that of hydrogen. Write the maximum and minimum possible kinetic energy of deuterium after collision.
Answer» ANSWER :`20 EV, 13.2 eV`

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