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:

The energy of the decay process can be calculated using:

= 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


= -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 :



= 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>
<em>I</em><em> </em><em>do</em><em> </em><em>not</em><em> </em><em>understand</em><em> </em><em>science</em><em> </em><em>but</em><em> </em><em>if</em><em> </em><em>u</em><em> </em><em>ask</em><em> </em><em>me</em><em> </em><em>I</em><em> </em><em>would</em><em> </em><em>have</em><em> </em><em>no</em><em> </em><em>clue</em><em> </em><em>do</em><em> </em><em>u</em><em> </em><em>get</em><em> </em><em>what</em><em> </em><em>I</em><em> </em><em>mean</em>
I believe the answer is D
The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions.