Question

Write the nuclear reaction equations for:
A.$\alpha -$decay of $_{88}^{226}Ra$
B.$\alpha -$decay of $_{94}^{242}Pu$
C.$\beta -$decay of $_{15}^{32}P$
D.$\beta -$decay of $_{83}^{210}Bi$
E.${\beta ^ + } -$decay of $_6^{11}C$
F.${\beta ^ + } -$decay of $_{43}^{97}Tc$
Electron capture of $_{54}^{120}Xe$

Answer 1

We have to know that in the case of $\alpha -$emission, there is an emission of an alpha particle and there is a formation of daughter nuclei. In the case of electron capture, there is no change in the number of masses and there is a formation of neutral atoms by simple electron capture. The gamma emission always follows the emission of alpha and beta decay. And a daughter nucleus is produced in the case of beta decay.

Complete answer:
(i) $\alpha -$decay of $_{88}^{226}Ra$ : By the alpha decay of radon, there is a formation of $_{86}^{222}Rn$ with helium and the reaction is,
$_{88}^{226}Rn \to _{86}^{222}Rn + _2^4He$
(ii) $\alpha -$decay of $_{94}^{242}Pu$: Plutonium $- 242$ is decayed into uranium $- 238$ along with helium. And the reaction can be written as,
$_{94}^{242}Pu \to _{92}^{238}U + _2^4He$
(iii) $\beta -$decay of $_{15}^{32}P$: The phosphorous $- 15$ undergoes negative beta emission without gain of any proton and there is an emission of antineutrino from the nucleus. The reaction is,
$_{15}^{32}P \to _{16}^{32}S + {e^ + } + {v^ - }$
(iv) $\beta -$decay of $_{83}^{210}Bi$: The bismuth $- 83$undergoes negative beta emission without gain of any proton and there is an emission of antineutrino from the nucleus. The reaction is,
$_{83}^{210}Bi \to _{84}^{210}Po + {e^ + } + {v^ - }$
(v) ${\beta ^ + } -$decay of $_6^{11}C$: Here the carbon $- 12$ undergoes the positive beta emission without gain of any proton and there is an emission of neutrino from the nucleus. The reaction is,
$_6^{11}C \to _5^{11}B + {e^ + } + v$
(vi) ${\beta ^ + } -$decay of $_{43}^{97}Tc$: Here the technetium $- 97$ undergoes the positive beta emission without gain of any proton and there is an emission of neutrino from the nucleus. The reaction is,
$_{43}^{97}Tc \to _{42}^{97}Mo + {e^ + } + v$
(vii) Electron capture of $_{54}^{120}Xe$ : Here, xenon $- 54$ undergoes electron capture and the reaction is,
$_{54}^{120}Xe + {e^ + } \to _{53}^{120}I + v$

Note:
We must have to know that the beta emission happens by the presence of too many protons in a nucleus and in that case, one of the neutrons or protons is converted into another. And here, the ratio of neutrons to protons is very high. Therefore, the remaining neutron is converted to an electron and proton. And the beta emission reaction is exothermic due to the formation of some amount of energy.