Question

The electron in a hydrogen atom at rest makes a transition from $n=2$ energy state to the $n=1$ ground state. Assume that all of the energy corresponding to transition from $n=2$ to $n=1$ is carried off by the photon. By setting the momentum of the system (atom+photon) equal to zero after the emission and assuming that the recoil energy of the atom is smaller compared with the $n=2$ to $n=1$ energy level separation, find the energy of the recoiling hydrogen atom.

• A. $2.75\times {10}^{-7}eV$
• B. $5.54\times {10}^{-8}eV$
• C. $8.11\times {10}^{-8}eV$
• D. $10.36\times {10}^{-7}eV$

Answer 1

The correct option is B $5.54\times {10}^{-8}eV$

Potential energy of the photon released is $=\frac{1}{2}(\frac{3}{4}\times 13.6\times 1.602\times {10}^{-19})=\frac{p\times 3\times {10}^8}{2}\Rightarrow p=5.46\times {10}^{-27}$ is the momentum of photon which is also equal to hydrogen.

Mass of hydrogen is, $m_H=1.67\times {10}^{-27}kg$

Therefore, recoil velocity is, $v=\frac{5.46\times {10}^{-27}}{1.67\times {10}^{-27}}=3.27m/s$

Energy of the atom is $=\frac{1}{2}(1.67\times {10}^{-27}){\left(3.27\right)}^2$$=8.92\times {10}^{-27}J$

which is $=\frac{8.92\times {10}^{-27}}{1.602\times {10}^{-19}}eV$

Therefore, Energy of hydrogen atom is $5.55\times {10}^{-8}eV$