Atomic recoil

Let us consider an atom or nucleus that emits a particle (a proton, neutron, alpha particle, neutrino, or gamma ray). In the simplest situation, the nucleus recoils with the same momentum, p as the particle has. The total energy of the "daughter" nucleus afterwards is

${\displaystyle {\sqrt {M_{d}^{2}c^{4}+p^{2}c^{2}}}}$

whereas that of the emitted particle is

${\displaystyle {\sqrt {M_{p}^{2}c^{4}+p^{2}c^{2}}}}$

where ${\displaystyle M_{d}}$ and ${\displaystyle M_{p}}$ are the rest masses of the daughter nucleus and the particle respectively. The sum of these must equal the rest energy of the original nucleus:

${\displaystyle {\sqrt {M_{p}^{2}c^{4}+p^{2}c^{2}}}+{\sqrt {M_{d}^{2}c^{4}+p^{2}c^{2}}}=M_{o}c^{2}}$

or

${\displaystyle {\sqrt {M_{p}^{2}c^{2}+p^{2}}}=M_{o}c-{\sqrt {M_{d}^{2}c^{2}+p^{2}}}.}$

Squaring both sides gives:

${\displaystyle M_{p}^{2}c^{2}+p^{2}=M_{o}^{2}c^{2}+M_{d}^{2}c^{2}+p^{2}-2M_{o}c{\sqrt {M_{d}^{2}c^{2}+p^{2}}}}$

or

${\displaystyle 2M_{o}c{\sqrt {M_{d}^{2}c^{2}+p^{2}}}=(M_{o}^{2}+M_{d}^{2}-M_{p}^{2})c^{2}.}$

Again squaring both sides gives:

${\displaystyle 4M_{o}^{2}c^{2}(M_{d}^{2}c^{2}+p^{2})=(M_{o}^{4}+M_{d}^{4}+M_{p}^{4}+2M_{o}^{2}M_{d}^{2}-2M_{o}^{2}M_{p}^{2}-2M_{d}^{2}M_{p}^{2})c^{4}}$

or

${\displaystyle p^{2}={\frac {M_{o}^{4}+M_{d}^{4}+M_{p}^{4}-2M_{o}^{2}M_{d}^{2}-2M_{o}^{2}M_{p}^{2}-2M_{d}^{2}M_{p}^{2}}{4M_{o}^{2}}}c^{2}}$

or

${\displaystyle p^{2}={\frac {(M_{o}-M_{d}-M_{p})(M_{o}+M_{d}-M_{p})(M_{o}-M_{d}+M_{p})(M_{o}+M_{d}+M_{p})}{4M_{o}^{2}}}c^{2}.}$

Note that ${\displaystyle (M_{o}-M_{d}-M_{p})c^{2}}$ is the energy released by the decay, which we may designate ${\displaystyle E_{decay}}$.

For the total energy of the particle we have:

${\displaystyle {\begin{aligned}{\sqrt {M_{p}^{2}c^{4}+p^{2}c^{2}}}&={\sqrt {{\frac {M_{o}^{4}+M_{d}^{4}+M_{p}^{4}-2M_{o}^{2}M_{d}^{2}+2M_{o}^{2}M_{p}^{2}-2M_{d}^{2}M_{p}^{2}}{4M_{o}^{2}}}c^{4}}}\\&={\frac {M_{o}^{2}-M_{d}^{2}+M_{p}^{2}}{2M_{o}}}c^{2}\\&=M_{p}c^{2}+{\frac {M_{o}^{2}-2M_{o}M_{p}+M_{p}^{2}-M_{d}^{2}}{2M_{o}}}c^{2}\\&=M_{p}c^{2}+{\frac {(M_{o}-M_{p})^{2}-M_{d}^{2}}{2M_{o}}}c^{2}\\&=M_{p}c^{2}+{\frac {1}{2}}\left({\frac {M_{o}-M_{p}}{M_{o}}}+{\frac {M_{d}}{M_{o}}}\right)(M_{o}-M_{d}-M_{p})c^{2}\end{aligned}}}$

So the kinetic energy imparted to the particle is:

${\displaystyle {\frac {1}{2}}\left({\frac {M_{o}-M_{p}}{M_{o}}}+{\frac {M_{d}}{M_{o}}}\right)E_{decay}}$

Similarly, the kinetic energy imparted to the daughter nucleus is:

${\displaystyle {\frac {1}{2}}\left({\frac {M_{o}-M_{d}}{M_{o}}}+{\frac {M_{p}}{M_{o}}}\right)E_{decay}}$

When the emitted particle is a proton, neutron, or alpha particle the fraction of the decay energy going to the particle is approximately ${\displaystyle M_{d}/M_{o}}$ and the fraction going to the daughter nucleus ${\displaystyle M_{p}/M_{o}.}$[3] For neutrinos and gamma rays, the departing particle gets almost all the energy, the fraction going to the daughter nucleus being only ${\displaystyle E_{decay}/(2M_{o}c^{2}).}$

The speed of the emitted particle is given by ${\displaystyle pc^{2}}$ divided by the total energy:

${\displaystyle v_{p}={\frac {\sqrt {(M_{o}-M_{d}-M_{p})(M_{o}+M_{d}-M_{p})(M_{o}-M_{d}+M_{p})(M_{o}+M_{d}+M_{p})}}{M_{o}^{2}-M_{d}^{2}+M_{p}^{2}}}c}$

Similarly, the speed of the recoiling nucleus is:

${\displaystyle v_{d}={\frac {\sqrt {(M_{o}-M_{d}-M_{p})(M_{o}+M_{d}-M_{p})(M_{o}-M_{d}+M_{p})(M_{o}+M_{d}+M_{p})}}{M_{o}^{2}+M_{d}^{2}-M_{p}^{2}}}c}$

When ${\displaystyle M_{p}=0}$ as for neutrinos and gamma rays, this simplifies to:

${\displaystyle {\begin{aligned}v_{d}&={\frac {M_{o}^{2}-M_{d}^{2}}{M_{o}^{2}+M_{d}^{2}}}c\\&={\frac {M_{o}+M_{d}}{M_{o}^{2}+M_{d}^{2}}}{\frac {E_{decay}}{c}}\\&\approx {\frac {2}{M_{o}+M_{d}}}{\frac {E_{decay}}{c}}\end{aligned}}}$

For similar decay energies, the recoil from emitting an alpha ray will be much greater than the recoil from emitting a neutrino (upon electron capture) or a gamma ray.

• nuclear recoil Britannica Online Encyclopedia