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#### Under certain circumstances, a nucleus can decay by emitting a particle more massive than an \alpha - particle. Consider the following decay processes: ^{223}_{88} Ra \rightarrow ^{209}_{82} Pb + ^{14}_{6} C

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^{223}_{88} Ra \rightarrow ^{219}_{86} Rn + ^{4}_{2} He .

Calculate the Q-values for these decays and determine that both are energetically allowed. Qalaxia Master Bot
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I found an answer from ncbgudi.com

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The amount of energy consumed or emitted during a nuclear reaction is known as the Q value. I can be determined from the difference between the masses of the initial reactants and the masses of the final products of a nuclear reaction. (MeV is a unit of energy).

Nuclear reactions

^{223}_{88} Ra \rightarrow ^{209}_{82} Pb + ^{14}_{6} C

^{223}_{88} Ra \rightarrow ^{219}_{86} Rn + ^{4}_{2} He

Step 1: Calculating the Q - value of first nuclear reactions

Mass of ^{223}_{88} Ra = 223.0183 u

Mass of ^{209}_{82} Pb = 208.98107 u

Mass of ^{14}_{6} C = 14.00324 u

\text{ Q -value } = \Delta m c^2   where, \Delta m - mass defect and c - speed of light

\Delta m = \text{ mass of } ^{223}_{88} Ra - \text{ mass of } ^{209}_{82} Pb - \text{ mass of } ^{14}_{6} C

\Delta m = 223.0183 u - 208.98107 - 14.00324 u

\Delta m = 0.03149 u

Q-value = 0.03149 c^2 u

Q-value = 0.03149 c^2 * 931.5 MeV/c^2          \because 1 u = 931.5 MeV/c^2

Q-value = 31.85 MeV

The reaction is (exothermic)energetically allowed  since the Q-value is positive.

Step 2: Calculating the Q - value of second nuclear reactions

Mass of ^{223}_{88} Ra = 223.0183 u

Mass of ^{219}_{86} Rn   = 219.00948

Mass of ^{4}_{2} He = 4.0026

Q - value $= [223.0183 u - 219.00948 - 4.0026] c^2$

Q - value = 0.00642 c^2 * * 931.5 MeV/c^2

Q- value = 5.98 MeV

The reaction is (exothermic)energetically allowed since the Q-value is positive.