Question

The emission of a photon by an isolated atom is a recoil process in which momentum is conserved. Thus Eq. (39.5) should include a recoil kinetic energy Kr for the atom. Why is this energy negligible in that equation?

Short answer

The de Broglie wavelength is 0.170nm, which is far too low compared to the required accuracy of$\pm$ 1.0nm.The wave nature need not to be considered.

Step-by-step solution

Important Concepts

The de Broglie wavelength is given by

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\sqrt{\mathbf{2}\mathbf{m}\mathbf{K}}}$

Where, m is the mass of the particle and K is the kinetic energy of the particle.

Replace the kinetic energy by$\mathit{K}\mathbf{=}\mathit{e}{\mathit{V}}_{\mathbf{b}\mathbf{a}}$ We get

$\mathbf{\lambda }\mathbf{=}\frac{\mathbf{h}}{\sqrt{\mathbf{2}\mathbf{m}\mathbf{2}{\mathbf{V}}_{\mathbf{ba}}}}$

Application

Input the values and we get

$$\begin{gathered} \lambda =\frac{6.626\times {10}^{-34}Js}{\sqrt{2\left(9.109\times {10}^{31}kg\right)\left(1.602\times {10}^{-19}C\right)\left(54V\right)}} \\ \lambda =0.170nm \end{gathered}$$

This is far too law compared to the required accuracy of $\pm 1.0nm$, so do not consider the wave nature.