91Ó°ÊÓ

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,

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

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

Substitute values in the above,

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

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 $1.67×{10}^{-27}\mathrm{kg}$.

Substitute all the values in the above,

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

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 $1.67×{10}^{-27}\mathrm{kg}$.

Substitute all values in the above,

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

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