A radioactive element {}_{92}^{242}X emits tw | vedclass

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Question

(A) The initial nucleus is ${}_{92}^{242}X$.An $\alpha$-particle is ${}_{2}^{4}He$,an electron $(\beta^-)$ is ${}_{-1}^{0}e$,and a positron $(\beta^+)

Vedclass · Accepted Answer

$87$

Vedclass · Answer

(A) The initial nucleus is ${}_{92}^{242}X$.An $\alpha$-particle is ${}_{2}^{4}He$,an electron $(\beta^-)$ is ${}_{-1}^{0}e$,and a positron $(\beta^+)$ is ${}_{+1}^{0}e$.The emission process is: ${}_{92}^{242}X \rightarrow 2({}_{2}^{4}He) + 1({}_{-1}^{0}e) + 2({}_{+1}^{0}e) + {}_{P}^{234}Y$.Conservation of atomic number $(Z)$: $92 = 2(2) + 1(-1) + 2(1) + P$.$92 = 4 - 1 + 2 + P$.$92 = 5 + P$.$P = 92 - 5 = 87$.

$A$ radioactive element ${}_{92}^{242}X$ emits two $\alpha$-particles,one electron,and two positrons. The product nucleus is represented by ${}_{P}^{234}Y$. The value of $P$ is $..................$

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