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

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

Vedclass · Accepted Answer

$87$

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(A) The initial nucleus is ${ }_{92}^{242} X$.An $\alpha$ particle is ${ }_{2}^{4} He$,an electron (beta-minus) is ${ }_{-1}^{0} e$,and a positron (beta-plus) is ${ }_{1}^{0} e$.The emission process is given by: ${ }_{92}^{242} X \rightarrow 2({ }_{2}^{4} He) + 1({ }_{-1}^{0} e) + 2({ }_{1}^{0} e) + { }_{P}^{234} Y$.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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