A hydrogen atom is at rest, in the first excited state, when it emits a photon of energy 10.2 eV. (a) What is the speed of the ground-state hydrogen atom when it recoils due to the photon emission? Remember that the magnitude of the momentum of a photon of energy E is p = E/c. Make the initial assumption that the kinetic energy of the recoiling atom is negligible compared to the photon energy. (b) Calculate the kinetic energy of the recoiling atom. Is this kinetic energy indeed negligible compared to the photon energy?

The given data can be listed below,

• The energy of photon is, $\mathrm{E}=10.2\mathrm{eV}$.
• The magnitude of momentum of photon is, $p=\frac{E}{c}$.

A dipole interaction causes photon emission by the hydrogen atom. the incident e-m field interacts with hydrogen's oscillating dipole moment. At the same time, it gathers and emits energy.

The momentum of the emitted photon is given by,

${\rho }_{\mathrm{photon}}=\frac{E}{c}$

Here, E is the energy of the photon and c is the speed of light

Substitute values in the above,

$$\begin{gathered} {\rho }_{\mathrm{photon}}=\frac{\left(10.2\right)\left({\displaystyle \frac{1.6\times {10}^{-19}\mathrm{J}}{1\mathrm{ev}}}\right)}{3.0\times {10}^8\mathrm{m}/\mathrm{s}}\left(\frac{1.1\times {10}^{-17}\mathrm{kg}}{1\mathrm{J}}\right) \\ =5.44\times {10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The speed of recoil hydrogen atom is given by,

$v_H=\frac{p_H}{m_H}$

Here, $p_H$is the momentum of the hydrogen which is equal to photon momentum, and is the mass of the hydrogen atom whose value is $1.67\times {10}^{-27}\mathrm{kg}$.

Substitute all the values in the above,

${\mathrm{v}}_{\mathrm{H}}=\frac{5.44\times {10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s}}{1.67\times {10}^{-27}\mathrm{kg}}$

Thus, the speed of recoiled hydrogen atom is 3.3 m/s.

The kinetic energy of the recoiling hydrogen atom is given by,

$K.E=\frac{{p^2}_H}{2m_H}$

Here, $p_H$is the momentum of the hydrogen, and is the mass of the hydrogen atom whose value is $1.67\times {10}^{-27}\mathrm{kg}$.

Substitute all values in the above,

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\frac{{\left(5.44\times {10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2}{1.67\times {10}^{-27}\mathrm{kg}} \\ =\left(8.84\times {10}^{-27}\mathrm{J}\right)\left(\frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right) \\ =5.5\times {10}^{-8}\mathrm{eV} \end{gathered}$$

Thus, the kinetic energy of the recoiling atom is$5.5\times {10}^{-8}\mathrm{eV}$ .