Question

The radionuclide${}^{\mathbf{32}}\mathbf{P}$decays to${}^{\mathbf{32}}\mathit{S}$as described by Eq. 42-24. In a particular decay event, an1.71 MeVelectron is emitted, the maximum possible value. What is the kinetic energy of the recoiling${}^{\mathbf{32}}\mathit{S}$atom in this event? (Hint:For the electron it is necessary to use the relativistic expressions for kinetic energy and linear momentum. The${}^{\mathbf{32}}\mathit{S}$atom is non-relativistic.)

Short answer

The kinetic energy of the recoiling ${}^{32}S$ atom in this event is 78.3 eV.

Step-by-step solution

The given data

The radionuclide decays${}^{32}P$ to ${}^{32}S$.

The maximum possible value of the emitted energy of the electron,$K_e=1.71MeV$

Understanding the concept of kinetic energy  

The kinetic energy of a particle in a decay process can be given as the maximum binding energy of the disintegrating process. Here, the decay of the radionuclide is beta decay releasing the energy due to the electron emission.

Formulae:

The kinetic energy of a particle, ${\mathrm{K}}_{max}=∆m{c}^2$ (1)

The relativistic relation of momentum-energy, ${\left(\mathrm{pc}\right)}^2={\mathrm{K}}^2+2{\mathrm{Kmc}}^2$ (2)

Calculation of the kinetic energy of the nuclide      

The beta decay of${}^{32}P$is given by:

${}^{32}P\underset{}{\to }{}^{32}S+e^{-}+\overline{v}$

However, since the electron has the maximum possible kinetic energy, no (anti)neutrino is emitted.

Since momentum is conserved, the momentum of the electron and the momentum of the residual sulfur nucleus are equal in magnitude and opposite in direction.

Ifis the momentum of the electron and$p_s$is the momentum of the sulfur nucleus, then using the conservation law of momentum$p_s=-p_e\dots \dots \dots ......\left(3\right)$.

The kinetic energy$K_s$of the sulfur nucleus is given using equation (1) and the value of the condition (3) as follows:

$$\begin{gathered} K_s=\frac{p^2}{2m_s} \\ =\frac{p^2}{2m_s}.............................\left(4\right) \end{gathered}$$

where,$m_s$is the mass of the sulfur nucleus. Now, the electron’s kinetic energy$K_e$is related to its momentum by the relativistic equation that is found using equation (2) as follows:${\left(p_{e}c\right)}^2=K_e^2+2K_{e}m{c}^2..................\left(5\right)$

wheremis the mass of an electron.

Thus, substituting the value of equation (b) in equation (a), we can get the kinetic energy of the recoiling sulfur nucleus using the given data as follows:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{s}}=\frac{{\mathrm{p}}_{\mathrm{e}}^2{\mathrm{c}}^2}{2{\mathrm{m}}_{\mathrm{s}}{\mathrm{c}}^2} \\ =\frac{{\mathrm{K}}_{\mathrm{e}}^2+2{\mathrm{K}}_{\mathrm{e}}{\mathrm{mc}}^2}{2{\mathrm{m}}_{\mathrm{s}}{\mathrm{c}}^2} \\ =\frac{{\left(1.71\mathrm{MeV}\right)}^2+2\left(1.71\mathrm{MeV}\right)\left(0.511\mathrm{MeV}\right)}{2\left(32\mathrm{u}\right)\left(931.5\mathrm{MeV}/\mathrm{u}\right)} \\ =7.83\times {10}^{-5}\mathrm{MeV} \\ =78.3\mathrm{eV} \end{gathered}$$

Hence, the kinetic energy of the recoiling sulfur nucleus is 78.3 eV.