Question

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $\alpha$-particle. Consider the following decay processes:
${}_{88}^{223}Ra\to {}_{82}^{209}Pb+{}_6^{14}C$
${}_{88}^{223}Ra\to {}_{86}^{219}Rn+{}_2^{4}He$
Calculate the Q-values for these decays and determine that both are energetically allowed.

Answer 1

(i) In case of first process,

${}_{88}^{223}Ra\to {}^{209}82Pb+{}_6^{14}C+Q$

The mass defect occured in reaction is,

$\Delta m=massofR{a}^{223}-(massofP{b}^{209}+massof{C}^{14})$

$=223.01850-(208.98107+14.00324)$

$=0.03419u$

So, the amount of energy released is given by:

$Q=\Delta m\times 931MeV$

$Q=0.03419\times 931MeV=31.83MeV$

(ii) Now, in case of second reaction,

${}_{88}^{223}Ra\to {}_{86}^{219}Rn+{}_2^{4}He+Q$

Mass defect is given as:

$\Delta m=massofR{a}^{223}-(massofR{n}^{219}+massofH{e}^4)$

$=223.01850-(219.00948+4.00260)$

$=0.00642u$

$\therefore$ the Energy released in the reaction is:

$Q=0.00642\times 931MeV=5.98MeV$

Since both reactions have positive Q-value so both reactions are possible.