Question

1. according to the equation:

$x \to \underset{82}{\overset{208}{\mathrm{pb}}}+\underset{2}{\overset{4}{\mathrm{he}}}$

the nucleus correctly represented by $x$ is

a) $\underset{80}{\overset{204}{\mathrm{hg}}}$ b) $\underset{84}{\overset{212}{\mathrm{po}}}$ c) $\underset{80}{\overset{204}{\mathrm{bi}}}$ d) $\underset{84}{\overset{212}{\mathrm{pb}}}$

1. when the nucleus of an atom of neon - 19 decays,

which particle is emitted?

a) $\underset{2}{\overset{4}{\mathrm{he}}}$ b) $\underset{-1}{\overset{0}{\mathrm{e}}}$ c) $\underset{0}{\overset{1}{\mathrm{n}}}$ d) $\underset{+1}{\overset{0}{\mathrm{e}}}$

Explanation:

Step1: Analyze the mass number

In a nuclear reaction, the sum of mass numbers on both sides must be equal.
For the reaction \(X
ightarrow_{82}^{208}Pb + _{2}^{4}He\), let the mass number of \(X\) be \(A\). Then \(A=208 + 4=212\)

Step2: Analyze the atomic number

The sum of atomic numbers on both sides must be equal. Let the atomic number of \(X\) be \(Z\). Then \(Z=82+2 = 84\)

for question 13:
Neon - 19 has atomic number \(Z = 10\) and mass number \(A=19\).
The general form of a nuclear decay reaction is \(_{10}^{19}Ne
ightarrow_{Z}^{A}Y+_{+1}^{0}e\) (where \(_{+1}^{0}e\) is a positron).
Using the conservation of atomic number: \(10=Z + 1\Rightarrow Z = 9\) (fluorine) and conservation of mass number \(19=A+0\Rightarrow A = 19\).
The reaction is \(_{10}^{19}Ne
ightarrow_{9}^{19}F+_{+1}^{0}e\)

Answer:

B. \(_{84}^{212}Po\)