A radioactive nucleus emits 4 alpha particles | vedclass

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(B) Let the initial nucleus be $^A_Z X$.Emission of one $\alpha$ particle $(^4_2 He)$ decreases the mass number by $4$ and atomic number by $2$.Emissi

Vedclass · Accepted Answer

$\frac{A-Z-15}{Z-1}$

Vedclass · Answer

(B) Let the initial nucleus be $^A_Z X$.Emission of one $\alpha$ particle $(^4_2 He)$ decreases the mass number by $4$ and atomic number by $2$.Emission of $4 \alpha$ particles results in a change: $A' = A - (4 	imes 4) = A - 16$ and $Z' = Z - (4 	imes 2) = Z - 8$.Emission of one $\beta$ particle $(^0_{-1} e)$ increases the atomic number by $1$ and leaves the mass number unchanged.Emission of $7 \beta$ particles results in: $A_{final} = A - 16$ and $Z_{final} = Z - 8 + 7 = Z - 1$.The number of protons in the final nucleus is $P = Z_{final} = Z - 1$.The number of neutrons is $N = A_{final} - Z_{final} = (A - 16) - (Z - 1) = A - Z - 15$.The ratio of neutrons to protons is $\frac{N}{P} = \frac{A - Z - 15}{Z - 1}$.

$A$ radioactive nucleus emits $4 \alpha$ particles and $7 \beta$ particles in succession. The ratio of the number of neutrons to that of protons in the final nucleus is $[A = \text{mass number}, Z = \text{atomic number}]$

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