
After $1\alpha $ and $2\beta $ emissions
A. Mass number reduces by $3$
B. Mass number reduces by $4$
C. Mass number reduces by $6$
D. Atomic number remains unchanged
Answer
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Hint:In order to solve this question, we should know that during radioactive process decays such as Alpha decay and beta decay occurs which result in change in atomic number and atomic mass of parent element, here firstly we will apply process of alpha decay on a general element and then beta decay two times and then we will determine change in mass number and atomic number.
Formula used:
General element is written as ${}_Z{X^A}$
where X is an element, Z is atomic number and A is atomic mass.
Alpha decay is just the release of a helium atom as ${}_2H{e^4}$ while beta decay just increases atomic number by one.
Complete step by step solution:
Let us suppose a general element ${}_Z{X^A}$ whose atomic number is Z and atomic mass is A and now according to the question, we have given that firstly one alpha decay occurs and as we know an alpha decay result in change in atomic number by $2$ and atomic mass by $4$ and this whole process can be shown as,
${}_Z{X^A}\mathop \to \limits^{1\alpha - decay} {}_{Z - 2}{X^{A - 4}}$
Now after forming the element ${}_{Z - 2}{X^{A - 4}}$ two beta decays will occur and as we know that one beta decay result in increase in atomic number by one so two beta decay will result in increase in atomic number by two and this process can be shown as,
${}_Z{X^A}\mathop \to \limits^{1\alpha - decay} {}_{Z - 2}{X^{A - 4}}\mathop \to \limits^{2\beta - decay} {}_Z{X^{A - 4}}$
So, the final element formed is ${}_Z{X^{A - 4}}$ where we see that the atomic number remains unchanged as Z and atomic mass is reduced by $4$ as $A \to A - 4$.
Hence, the correct answer is option B.
Note: While solving such questions, always remember the standard notation of an element and also don’t confuse with atomic number and atomic mass, atomic number is always written in subscript while atomic mass is written in superscript and remember the most known decays process such as alpha, beta and gamma decays.
Formula used:
General element is written as ${}_Z{X^A}$
where X is an element, Z is atomic number and A is atomic mass.
Alpha decay is just the release of a helium atom as ${}_2H{e^4}$ while beta decay just increases atomic number by one.
Complete step by step solution:
Let us suppose a general element ${}_Z{X^A}$ whose atomic number is Z and atomic mass is A and now according to the question, we have given that firstly one alpha decay occurs and as we know an alpha decay result in change in atomic number by $2$ and atomic mass by $4$ and this whole process can be shown as,
${}_Z{X^A}\mathop \to \limits^{1\alpha - decay} {}_{Z - 2}{X^{A - 4}}$
Now after forming the element ${}_{Z - 2}{X^{A - 4}}$ two beta decays will occur and as we know that one beta decay result in increase in atomic number by one so two beta decay will result in increase in atomic number by two and this process can be shown as,
${}_Z{X^A}\mathop \to \limits^{1\alpha - decay} {}_{Z - 2}{X^{A - 4}}\mathop \to \limits^{2\beta - decay} {}_Z{X^{A - 4}}$
So, the final element formed is ${}_Z{X^{A - 4}}$ where we see that the atomic number remains unchanged as Z and atomic mass is reduced by $4$ as $A \to A - 4$.
Hence, the correct answer is option B.
Note: While solving such questions, always remember the standard notation of an element and also don’t confuse with atomic number and atomic mass, atomic number is always written in subscript while atomic mass is written in superscript and remember the most known decays process such as alpha, beta and gamma decays.
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