A nucleus _Z{X^A} emits 9 alpha-particles and | vedclass

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Question

(A) The initial nucleus is $_Z{X^A}$.When it emits $9 \alpha$-particles,the atomic number $Z$ decreases by $9 	imes 2 = 18$ and the mass number $A$ d

Vedclass · Accepted Answer

$\frac{Z - 13}{A - Z - 23}$

Vedclass · Answer

(A) The initial nucleus is $_Z{X^A}$.When it emits $9 \alpha$-particles,the atomic number $Z$ decreases by $9 	imes 2 = 18$ and the mass number $A$ decreases by $9 	imes 4 = 36$. The nucleus becomes $_{Z-18}{X^{A-36}}$.When this nucleus emits $5 \beta$-particles,the atomic number $Z$ increases by $5 	imes 1 = 5$,while the mass number $A$ remains unchanged. The final nucleus is $_{Z-18+5}{X^{A-36}} = _{Z-13}{X^{A-36}}$.In the final nucleus,the number of protons $P = Z - 13$.The number of neutrons $N = (A - 36) - (Z - 13) = A - Z - 23$.Therefore,the ratio of protons to neutrons is $\frac{P}{N} = \frac{Z - 13}{A - Z - 23}$.

$A$ nucleus $_Z{X^A}$ emits $9 \alpha$-particles and $5 \beta$-particles. The ratio of total protons to neutrons in the final nucleus is:

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