An element _{96}X^{227} emits 4alpha and 5bet | vedclass

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Question

(A) The nuclear reaction is given by: $_{96}X^{227} 	o Y + 4(_{2}He^{4}) + 5(_{-1}e^{0})$.To find the mass number $(A)$ of $Y$: $227 = A + 4 	imes 4

Vedclass · Accepted Answer

$93; 211$

Vedclass · Answer

(A) The nuclear reaction is given by: $_{96}X^{227} 	o Y + 4(_{2}He^{4}) + 5(_{-1}e^{0})$.To find the mass number $(A)$ of $Y$: $227 = A + 4 	imes 4 + 5 	imes 0 \implies 227 = A + 16 \implies A = 211$.To find the atomic number $(Z)$ of $Y$: $96 = Z + 4 	imes 2 + 5 	imes (-1) \implies 96 = Z + 8 - 5 \implies 96 = Z + 3 \implies Z = 93$.Thus,the atomic number is $93$ and the mass number is $211$.

An element $_{96}X^{227}$ emits $4\alpha$ and $5\beta$ particles to form a new element $Y$. The atomic number and mass number of $Y$ are:

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