
The ${}^{22}Ne$ nucleus, after absorbing energy, decays into two α-particles and an unknown nucleus. The unknown nucleus is
A. Nitrogen
B. Carbon
C. Boron
D. Oxygen
Answer
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Hint:In order to solve this question, we will first apply the process of an alpha decay during radioactivity and using the general notation for an element we will find the atomic number and atomic mass of the newly formed element due to two alpha decays and then determine the correct option.
Formula used:
During an alpha decay, a parental element atomic number gets reduced by two while atomic mass of the element gets reduced by four, and general notation of an element is ${}_Z{X^A}$ where X is an element, Z is atomic number and A is atomic mass.
Complete step by step solution:
According to the question, we have given that the given nucleus is ${}^{22}Ne$ which is Neon and whose atomic number is $10$, so the chemical element is written as ${}_{10}N{e^{22}}$ and as we know that during one alpha decay the element’s atomic number will get reduced by two while atomic mass gets reduced by four and this beta decay can be represented as,
\[{}_{10}N{e^{22}}\mathop \to \limits^{\alpha - decay} {}_{10 - 2}N{e^{22 - 4}} \\
\Rightarrow {}_{10}N{e^{22}}\mathop \to \limits^{\alpha - decay} {}_8{Y^{18}} \\
\]
where Y is some formed element and now, this element Y goes further one alpha decay process and this can be represented as,
\[{}_8{Y^{18}}\mathop \to \limits^{\alpha - decay} {}_{8 - 2}{P^{18 - 4}} \\
\Rightarrow {}_8{Y^{14}}\mathop \to \limits^{\alpha - decay} {}_6{P^{14}} \\ \]
So, the final chemical element formed is \[{}_6{P^{14}}\] whose atomic number is $6$ and atomic mass is \[14\], and this element is Carbon.
Hence, the correct answer is option B.
Note:While solving such problems, always make sure the proper notations of element and in standard form the atomic number is always written in left subscript while atomic mass in right superscript, and just remember that alpha decay is just release of doubly ionised helium atom to avoid any subtraction errors on atomic number and atomic mass.
Formula used:
During an alpha decay, a parental element atomic number gets reduced by two while atomic mass of the element gets reduced by four, and general notation of an element is ${}_Z{X^A}$ where X is an element, Z is atomic number and A is atomic mass.
Complete step by step solution:
According to the question, we have given that the given nucleus is ${}^{22}Ne$ which is Neon and whose atomic number is $10$, so the chemical element is written as ${}_{10}N{e^{22}}$ and as we know that during one alpha decay the element’s atomic number will get reduced by two while atomic mass gets reduced by four and this beta decay can be represented as,
\[{}_{10}N{e^{22}}\mathop \to \limits^{\alpha - decay} {}_{10 - 2}N{e^{22 - 4}} \\
\Rightarrow {}_{10}N{e^{22}}\mathop \to \limits^{\alpha - decay} {}_8{Y^{18}} \\
\]
where Y is some formed element and now, this element Y goes further one alpha decay process and this can be represented as,
\[{}_8{Y^{18}}\mathop \to \limits^{\alpha - decay} {}_{8 - 2}{P^{18 - 4}} \\
\Rightarrow {}_8{Y^{14}}\mathop \to \limits^{\alpha - decay} {}_6{P^{14}} \\ \]
So, the final chemical element formed is \[{}_6{P^{14}}\] whose atomic number is $6$ and atomic mass is \[14\], and this element is Carbon.
Hence, the correct answer is option B.
Note:While solving such problems, always make sure the proper notations of element and in standard form the atomic number is always written in left subscript while atomic mass in right superscript, and just remember that alpha decay is just release of doubly ionised helium atom to avoid any subtraction errors on atomic number and atomic mass.
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