13.1 Recoil speed of electron on emission of a photon

  Basic Modern Physics

    1.0 Photon theory of light
    2.0 Characteristics of photon
    3.0 Wave Particle Duality
    4.0 Emission of electrons
    5.0 Photoelectric Effect
    6.0 Radiation Pressure And Force
    7.0 Photon Density
    8.0 Force exerted by a light beam on a surface
    9.0 Early Atomic Structures
    10.0 Bohr Model of The Hydrogen Atom
    11.0 Energy of electron in the $n^{th}$ orbit
    12.0 Basic Definitions
    13.0 Atomic Excitation

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13.1 Recoil speed of electron on emission of a photon

As the atom is fixed, so, all the energy released during de-excitation is transferred to the emitted photon with no change in KE of the atom. Here, $$\begin{equation} \begin{aligned} \frac{{hc}}{\lambda } = \Delta {E_{\left( {{n_2} \to {n_1}} \right)}} \\ \\\end{aligned} \end{equation} $$

However, energy released during de-excitation is distributed between the atom and the released photon. By CLM, $$\begin{equation} \begin{aligned} mv = \frac{h}{{\lambda '}} \\ \\\end{aligned} \end{equation} $$ By CE, $$\begin{equation} \begin{aligned} \frac{1}{2}m{v^2} + \frac{{hc}}{{\lambda '}} = \Delta {E_{\left( {{n_2} \to {n_1}} \right)}} \\ \Rightarrow \frac{{hc}}{{\lambda '}} = \Delta {E_{\left( {{n_2} \to {n_1}} \right)}}{\text{ (neglecting KE of atom as it has very large mass than photon)}} \\ \Rightarrow \frac{h}{{\lambda '}} = \frac{{\Delta {E_{\left( {{n_2} \to {n_1}} \right)}}}}{c} \\ \Rightarrow mv = \frac{{\Delta {E_{\left( {{n_2} \to {n_1}} \right)}}}}{c} \\ {\text{or }}{v_{{\text{(recoil speed)}}}} = \frac{{\Delta {E_{\left( {{n_2} \to {n_1}} \right)}}{\text{ }}}}{{mc}}{\text{ }} \\\end{aligned} \end{equation} $$

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Question 15. A $H{e^ + }$ ion is at rest and is in ground state. A neutron with Initial kinetic energy $K$ collides head on with the $H{e^ + }$ ion. Find minimum value of $K$ so that there can be an inelastic collision between these two particles.

Solution: Here, the loss during the collision can only be used to excite the atoms or electrons.

So, according to quantum mechanics, possible losses can be ${0,\ 40.8\ eV,\ 48.36\ eV,...,\ 54.4\ eV }\ \ \ ...(i)$

$$\begin{equation} \begin{aligned} {\text{from }}{E_n} = - 13.6\frac{{{Z^2}}}{{{n^2}}}\ eV \\ \\\end{aligned} \end{equation} $$

Now, according to Newtonian mechanics: Minimum loss$=0$

Maximum loss will be for perfectly inelastic collision.

Let $v_0$ be the initial speed of neutron and $v_f$ be the final common speed.

So, by momentum conservation:

$$\begin{equation} \begin{aligned} m{v_0} = m{v_f} + 4m{v_f}\quad \Rightarrow \quad {v_f} = \frac{{{v_0}}}{5} \\ \\\end{aligned} \end{equation} $$

where $m=$ mass of neutron and mass of $H{e^ + }$ ion$=4m$

Final kinetic energy of system:

$$\begin{equation} \begin{aligned} KE = \frac{1}{2}mv_f^2 + \frac{1}{2}\left( {4mv_f^2} \right) \\ = \frac{1}{2}\left( {5m} \right)\frac{{v_0^2}}{{25}} = \frac{1}{5}\left( {\frac{1}{2}mv_0^2} \right) = \frac{K}{5} \\ {\text{Maximum loss}} = K - \frac{K}{5} = \frac{{4K}}{5} \\ {\text{So, loss will be }}\left[ {0,\frac{{4K}}{5}} \right]\quad \quad \quad ...(ii) \\\end{aligned} \end{equation} $$

For inelastic collision, there should be at least one common value other than zero in Equation. $(i)$ and $(ii)$

$$\begin{equation} \begin{aligned} \therefore \frac{{4K}}{5} > 40.8\ eV\quad \Rightarrow \quad K > 51\ eV \\ \\\end{aligned} \end{equation} $$

Hence, minimum value of $K$ for an inelastic collision: ${K_{\min }} = 51\ eV$