Ncert Exercises & Solutions

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

$\begin{aligned} &{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C} \\ &{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} \end{aligned}$

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

Hint: The Q-value is given by

$E=\Delta mc^2$ .
Step 1: Find the Q-value for first reaction.

$F o r t h e e m i s s i o n o f {}_{6}^{14}C, t h e n u c l e a r r e a c t i o n : {}_{88}^{223}R a \to {}_{82}^{209}P b + {}_{6}^{14}C$

$W e k n o w t h a t :$

$M a s s o f {}_{88}^{223}R a, m_{1} = 223.01850 u$

$M a s s o f {}_{82}^{209}P b, m_{2} = 208.98107 u$

$M a s s o f {}_{6}^{14}C, m_{3} = 14.00324 u$

$H e n c e, t h e Q - v a l u e o f t h e r e a c t i o n i s g i v e n a s :$

$Q = (m_{1} - m_{2} - m_{3}) c^{2} = (223.01850 - 208.98107 - 14.00324) c^{2} = (0.03419 c^{2}) u$

$B u t 1 u = 931.5 M e V / c^{2}$

$∴ Q = 0.03419 \times 931.5 = 31.848 M e V$

$H e n c e, t h e Q - v a l u e o f t h e n u c l e a r r e a c t i o n i s 31.848 M e V . S i n c e t h e v a l u e i s p o s i t i v e, t h e r e a c t i o n i s e n e r g e t i c a l l y a l l o w e d .$

$Step 2:Find the Q-value for the second reaction.$
For the emission of

${ }_{2}^{4} \mathrm{He}$ , the nuclear reaction is:

${ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He}$
We know that:
Mass of

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

${ }_{86}^{219} R n, m_{2}=219.00948~u$
Mass of

${ }_{2}^{4} \mathrm{He}, m_{3}=4.00260~u$

$Q - v a l u e o f t h i s n u c l e a r r e a c t i o n i s g i v e n a s :$

$Q = (m_{1} - m_{2} - m_{3}) c^{2} = (223.01850 - 219.00948 - 4.00260) C^{2} = (0.00642 c^{2}) u =$

$= 0.00642 \times 931.5 = 5.98 M e V$

Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetcally allowed.

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