Question 420: Atoms & Nuclei

Question

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

Accepted Answer

We are given:
A nucleus $\,^{242}_{92}X\,$.
It undergoes the following emissions:
Two $\alpha$-particles
(helium nuclei: $\,^{4}_{2}\mathrm{He}$):
Each $\alpha$ emission decreases the
mass number
$A$ by 4 and the
atomic number
$Z$ by 2.
So, two $\alpha$ particles:
Decrease in $A$: $2 	imes 4 = 8$.
Decrease in $Z$: $2 	imes 2 = 4$.
Thus after two $\alpha$ emissions:
$ A = 242 - 8 = 234,\quad Z = 92 - 4 = 88. $
Intermediate nucleus: $\,^{234}_{88}$.
One electron ($\beta^-$) emission
:
A $\beta^-$ is emission of an electron when a neutron Ã¢ÂÂ proton.
So
atomic number $Z$ increases by 1
, mass number unchanged.
After this:
$ A = 234,\quad Z = 88 + 1 = 89. $
Two positron ($\beta^+$) emissions
:
Each $\beta^+$ emission: proton Ã¢ÂÂ neutron, so
atomic number $Z$ decreases by 1
.
Two positrons Ã¢ÂÂ decrease $Z$ by 2.
After this:
$ A = 234,\quad Z = 89 - 2 = 87. $
Final nucleus: $\,^{234}_{87}Y$.
So, the value of $P$ is:
$ \boxed{87} $
Correct Option: A (87).
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Answer

87

Answer

85

Answer

92

Answer

96

Atoms & Nuclei

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

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