Answer:
a). The energy released by the alpha decay of 222 Rn is to 218 Po :
= 5.596 MeV
b).The energy of the alpha particle is 5.5 MeV
c) The recoil energy of Po = 0.096 MeV
Explanation:
The equation for the alpha decay of Rn to Polonium is:
![_{2}^{4}\textrm{He}+_{84}^{218}\textrm{Po}](https://tex.z-dn.net/?f=_%7B2%7D%5E%7B4%7D%5Ctextrm%7BHe%7D%2B_%7B84%7D%5E%7B218%7D%5Ctextrm%7BPo%7D)
The energy of the decay process can be calculated using:
![\Delta E=\Delta mc^{2}](https://tex.z-dn.net/?f=%5CDelta%20E%3D%5CDelta%20mc%5E%7B2%7D)
= Change in the mass
Mass of Po = 218.008965 u
Mass of He = 4.002603 u
Mass of Rn = 222.017576 u
= mass of Po + mass of He - mass of Rn
= 4.002603 + 218.008965-222.017576
= - 0.006008
![\Delta E=m(931.5MeV)](https://tex.z-dn.net/?f=%5CDelta%20E%3Dm%28931.5MeV%29)
![\Delta E=-0.006008\times 931.5MeV](https://tex.z-dn.net/?f=%5CDelta%20E%3D-0.006008%5Ctimes%20931.5MeV)
= -5.596 MeV
<u><em>The negative sign means energy is released during the process.</em></u>
b) The energy of the alpha particle is :
![\frac{Po\ mass }{Rn\ mass}\times E](https://tex.z-dn.net/?f=%5Cfrac%7BPo%5C%20mass%20%7D%7BRn%5C%20mass%7D%5Ctimes%20E)
![\frac{218.0089}{222.0175}\times \Delta E](https://tex.z-dn.net/?f=%5Cfrac%7B218.0089%7D%7B222.0175%7D%5Ctimes%20%5CDelta%20E)
![\frac{218.0089}{222.0175}\times 5.596](https://tex.z-dn.net/?f=%5Cfrac%7B218.0089%7D%7B222.0175%7D%5Ctimes%205.596)
= 5.494 MeV
= 5.5 MeV
c).
Recoil energy: When the parent nucleus is at rest before the decay then , there must be the some recoil of daughter nucleus to conserve the momentum.This is termed as recoil energy.
<u><em>The energy of the recoil polonium atom :</em></u>
The formula for recoil energy is :
<em>The total energy - the kinetic energy</em>
<em>= 5.596 - 5.5 </em>
<em>= 0.096 MeV</em>