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# Problem P. 4862. (September 2016)

P. 4862. When the nucleus of a radium-226, which is at rest, decays, an $\displaystyle \alpha$ particle and the nucleus of radon are created. The released energy during the decay is 4.87 MeV, which appears as the kinetic energy of the two resulted particles. Calculate the linear momentum and the kinetic energy of both the $\displaystyle \alpha$ particle and the radon nucleus.

(4 pont)

Deadline expired on October 10, 2016.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A $\displaystyle m$ tömegű $\displaystyle \alpha$-részecske és a $\displaystyle M$ tömegű radonmag lendülete ugyanakkora $\displaystyle p$ nagyságú, hiszen a rádium lendülete nulla volt. A mozgási energiák összege kifejezhető a lendülettel:

$\displaystyle p=mv,\qquad \frac{1}{2}mv^2=\frac{p^2}{2m},$

$\displaystyle \frac{p^2}{2m}+\frac{p^2}{2M}=\Delta E,$

ahonnan

$\displaystyle p= \sqrt{ 2\Delta E\, \frac{mM}{m+M}}.$

Tudjuk, hogy

$\displaystyle \Delta E=4{,}87~{\rm MeV}=7{,}8\cdot 10^{-13}~{\rm J},$

továbbá

$\displaystyle m=4{,}002~m_{\rm u}=6{,}64\cdot10^{-27}~{\rm kg},\qquad M=222{,}017~m_{\rm u}=3{,}685\cdot10^{-25}~{\rm kg}.$

A bomlástermékek lendülete: $\displaystyle p=1{,}01\cdot10^{-19}~{\rm kg}\frac{\rm m}{\rm s},$ a mozgási energiájuk pedig

$\displaystyle E_{\alpha\text{-rész}}=\frac{p^2}{2m}=\frac{M}{m+M}\Delta E=\frac{222}{4+222}\cdot4{,}87~{\rm MeV}=4{,}78~{\rm MeV},$

illetve

$\displaystyle E_{\text{radon}}=\frac{p^2}{2M}=\frac{m}{m+M}\Delta E=\frac{4}{4+222}\cdot4{,}87~{\rm MeV}=0{,}09~{\rm MeV}.$

Látható, hogy a felszabaduló energia legnagyobb részét a könnyű bomlástermék (az alfa-részecske) ,,viszi el''.

### Statistics:

 78 students sent a solution. 4 points: 53 students. 3 points: 11 students. 2 points: 5 students. 1 point: 5 students. 0 point: 4 students.

Problems in Physics of KöMaL, September 2016