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The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element $X$ as shown below. To which group, element $X$ belongs in the periodic table?
${ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6_{0}^{1} n+{ }_{2}^{4} \alpha+2{ }_{1}^{1} \mathrm{H}+\mathrm{X}$
${ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6_{0}^{1} n+{ }_{2}^{4} \alpha+2{ }_{1}^{1} \mathrm{H}+\mathrm{X}$
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Verified Answer
The correct answer is:
8
${ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6{ }_{0}^{1} n+{ }_{2}^{4} \mathrm{He}+2{ }_{1}^{1} \mathrm{H}+{ }_{Z}^{A} X$
Balancing the atomic mass and atomic number
$63+1=(6 \times 1)+4+2+A \Rightarrow A=52$
$29+1=(6 \times 0)+2+2+Z \Rightarrow Z=26$
Thus ${ }_{Z}^{A} X={ }_{26}^{52} X$ or ${ }_{26}^{52} \mathrm{Fe}$
Hence, $X$ belongs to group 8 in the periodic table.
Balancing the atomic mass and atomic number
$63+1=(6 \times 1)+4+2+A \Rightarrow A=52$
$29+1=(6 \times 0)+2+2+Z \Rightarrow Z=26$
Thus ${ }_{Z}^{A} X={ }_{26}^{52} X$ or ${ }_{26}^{52} \mathrm{Fe}$
Hence, $X$ belongs to group 8 in the periodic table.
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