Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-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.

Solution

(i) For the first decay process,

${}_{88}^{223}{Ra} \to {}_{82}^{209}{Pb} + {}_{6}^{14}C + Q$

Mass defect,

$∆ m = mass of {Ra}^{223} - (mass of {Pb}^{209} + mass of C^{14})$
$= 223.01850 - (208.98107 + 14.00324) = 0.03419 u$
Therefore,
Energy released, $Q = 0.03419 \times 931 MeV = 31.83 MeV$

(ii) Now, for second decay process we have,

${}_{88}^{223}{Ra} \to {}_{86}^{219}{Rn} + {}_{2}^{4}{He} + Q$

Mass defect, $∆ m$ = mass of ${Ra}^{223}$ - (mass of Rn²¹⁹ + mass of He⁴)

$= 223.01850 - (219.00948 + 4.00260) = 0.00642 u$

$∴$ Energy released, $Q = 0.00642 \times 931 MeV = 5.98 MeV$

Both the above given decay processes are energetically possible because the Q-value i.e., energy released in the reaction is positive.

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