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, {{mvar|p}} as the particle has. The total energy of the "daughter" nucleus afterwards is

:$\backslash sqrt\{M\_d^2c^4+p^2c^2\}$

whereas that of the emitted particle is

:$\backslash sqrt\{M\_p^2c^4+p^2c^2\}$

where $M\_d$ and $M\_p$ 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:

:$\backslash sqrt\{M\_p^2c^4+p^2c^2\}+\backslash sqrt\{M\_d^2c^4+p^2c^2\}=M\_oc^2$

or

:$\backslash sqrt\{M\_p^2c^2+p^2\}=M\_oc-\backslash sqrt\{M\_d^2c^2+p^2\}.$

Squaring both sides gives:

:$M\_p^2c^2+p^2=M\_o^2c^2+M\_d^2c^2+p^2-2M\_oc\backslash sqrt\{M\_d^2c^2+p^2\}$

or

:$2M\_oc\backslash sqrt\{M\_d^2c^2+p^2\}=(M\_o^2+M\_d^2-M\_p^2)c^2.$

Again squaring both sides gives:

:$4M\_o^2c^2(M\_d^2c^2+p^2)=(M\_o^4+M\_d^4+M\_p^4+2M\_o^2M\_d^2-2M\_o^2M\_p^2-2M\_d^2M\_p^2)c^4$

or

:$p^2=\backslash frac\{M\_o^4+M\_d^4+M\_p^4-2M\_o^2M\_d^2-2M\_o^2M\_p^2-2M\_d^2M\_p^2\}\{4M\_o^2\}c^2$

or

:$p^2=\backslash frac\{(M\_o-M\_d-M\_p)(M\_o+M\_d-M\_p)(M\_o-M\_d+M\_p)(M\_o+M\_d+M\_p)\}\{4M\_o^2\}c^2.$

Note that $(M\_o-M\_d-M\_p)c^2$ is the energy released by the decay, which we may designate $E\_\{decay\}$.

For the total energy of the particle we have:

:$\backslash begin\{align\}$

\sqrt{M_p^2c^4+p^2c^2}&=\sqrt{\frac{M_o^4+M_d^4+M_p^4-2M_o^2M_d^2+2M_o^2M_p^2-2M_d^2M_p^2}{4M_o^2}c^4}\\

&=\frac{M_o^2-M_d^2+M_p^2}{2M_o}c^2\\

&=M_pc^2+\frac{M_o^2-2M_oM_p+M_p^2-M_d^2}{2M_o}c^2\\

&=M_pc^2+\frac{(M_o-M_p)^2-M_d^2}{2M_o}c^2\\

&=M_pc^2+\frac 12\left(\frac{M_o-M_p}{M_o}+\frac{M_d}{M_o}\right)(M_o-M_d-M_p)c^2

\end{align}

So the kinetic energy imparted to the particle is:

:$\backslash frac\; 12\backslash left(\backslash frac\{M\_o-M\_p\}\{M\_o\}+\backslash frac\{M\_d\}\{M\_o\}\backslash right)E\_\{decay\}$

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

:$\backslash frac\; 12\backslash left(\backslash frac\{M\_o-M\_d\}\{M\_o\}+\backslash frac\{M\_p\}\{M\_o\}\backslash right)E\_\{decay\}$

When the emitted particle is a proton, neutron, or alpha particle the fraction of the decay energy going to the particle is approximately $M\_d/M\_o$ and the fraction going to the daughter nucleus $M\_p/M\_o.${{cite book

|title=Concepts of Modern Physics

|url=http://phy240.ahepl.org/Concepts_of_Modern_Physics_by_Beiser.pdf

|year=2003

|publisher=McGraw-Hill

|isbn=0-07-244848-2

|chapter=Chapter 12: Nuclear Transformations

|pages=432–434

|edition=6th

|author1=Arthur Beiser

|access-date=2016-07-03

|archive-url=https://web.archive.org/web/20161004204701/http://phy240.ahepl.org/Concepts_of_Modern_Physics_by_Beiser.pdf

|archive-date=2016-10-04

|url-status=dead

}}

For neutrinos and gamma rays, the departing particle gets almost all the energy, the fraction going to the daughter nucleus being only $E\_\{decay\}/(2M\_oc^2).$

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

:$v\_p=\backslash frac\{\backslash sqrt\{(M\_o-M\_d-M\_p)(M\_o+M\_d-M\_p)(M\_o-M\_d+M\_p)(M\_o+M\_d+M\_p)\}\}\{M\_o^2-M\_d^2+M\_p^2\}c$

Similarly, the speed of the recoiling nucleus is:

:$v\_d=\backslash frac\{\backslash sqrt\{(M\_o-M\_d-M\_p)(M\_o+M\_d-M\_p)(M\_o-M\_d+M\_p)(M\_o+M\_d+M\_p)\}\}\{M\_o^2+M\_d^2-M\_p^2\}c$

If we take $M\_p=0$ for neutrinos and gamma rays, this simplifies to:

:$\backslash begin\{align\}$

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{align}

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.

For decays that produce two particles as well as the daughter nuclide, the above formulas can be used to find the maximum energy, momentum, or speed of any of the three, by assuming that the lighter of the other two ends up with a speed of zero. For example, the maximum energy of the neutrino, if we assume its rest mass to be zero, is found by using the formula as though only the daughter and the neutrino are involved:

:$\backslash frac\; 12\backslash left(1+\backslash frac\{M\_d\}\{M\_o-M\_e\}\backslash right)E\_\{decay\}$

Note that $M\_d$ here is not the mass of the neutral daughter isotope, but that minus the electron mass: $M\_d=M\_o-E\_\{decay\}/c^2-M\_e.$

With beta decay, the maximum recoil energy of the daughter nuclide, as a fraction of the decay energy, is greater than either of the approximations given above, $M\_e/M\_o.$ and $E\_\{decay\}/(2M\_oc^2).$ The first ignores the decay energy, and the second ignores the mass of the beta particle, but with beta decay these two are often comparable and neither can be ignored (see Beta decay#Energy release).

{{Reflist}}

• {{cite book |last=Hahn |first=Otto |translator-first=Willy |translator-last=Ley |title=Otto Hahn: A Scientific Autobiography |year=1966 |publisher=Charles Scribner's Sons |location=New York |oclc=646422716}}
• {{cite book |last1=Gerlach |first1=Walther |author-link=Walther Gerlach |last2=Hahn |first2=Dietrich |title=Otto Hahn – Ein Forscherleben unserer Zeit |language=de |publisher=Wissenschaftliche Verlagsgesellschaft (WVG) |location=Stuttgart |year=1984 |isbn=978-3-8047-0757-3 |oclc=473315990}}

• [http://www.britannica.com/EBchecked/topic/421785/nuclear-recoil nuclear recoil Britannica Online Encyclopedia]

Category:Atomic physics