A radioactive nucleus (initial mass number A | vedclass

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Question

(B) The initial nucleus is represented as $_Z^AX$.
An $\alpha$-particle is $_2^4He$ and a positron is $_1^0e^+$.
After emitting $3$ $\alpha$-particl

Vedclass · Accepted Answer

$\frac{A - Z - 8}{Z - 4}$

Vedclass · Answer

(B) The initial nucleus is represented as $_Z^AX$.
An $\alpha$-particle is $_2^4He$ and a positron is $_1^0e^+$.
After emitting $3$ $\alpha$-particles and $2$ positrons, the final nucleus $_Z^AY$ is formed.
The new mass number $A' = A - (3 	imes 4) = A - 12$.
The new atomic number $Z' = Z - (3 	imes 2) + (2 	imes 1) = Z - 6 + 2 = Z - 4$.
The number of protons in the final nucleus is $P = Z' = Z - 4$.
The number of neutrons in the final nucleus is $N = A' - Z' = (A - 12) - (Z - 4) = A - Z - 8$.
The ratio of neutrons to protons is $\frac{N}{P} = \frac{A - Z - 8}{Z - 4}$.

$A$ radioactive nucleus (initial mass number $A$ and atomic number $Z$) emits $3$ $\alpha$-particles and $2$ positrons. The ratio of the number of neutrons to that of protons in the final nucleus will be:

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