Physics Ncert Solutions Class 12th Nuclei Class 12th Physics Ncert Solutions Physics Class 12th

13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $α -$ particle. Consider the following decay processes:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

Calculate the Q-values for these decays and determine that both are energetically allowed.

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Payal Gupta | Contributor-Level 10

7 months ago

13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 31.848 MeV

Hence,

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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