Question

What are the respective numbers of \(\alpha\) and \(\beta\) particles emitted respectively in the following radioactive decay? \[ {}^{200}_{90}X \rightarrow {}^{168}_{80}Y \]

• A. \(8\) and \(8\)
• B. \(8\) and \(6\)
• C. \(6\) and \(8\)
• D. \(6\) and \(6\)

Hint:

For \(\alpha\)-decay, mass number decreases by \(4\) and atomic number decreases by \(2\). For \(\beta^{-}\)-decay, mass number remains same and atomic number increases by \(1\).

Answer 1

The Correct Option is B

Concept:
In radioactive decay: \[ \alpha\text{-particle} = {}^{4}_{2}\text{He} \] So, emission of one \(\alpha\)-particle decreases mass number by \(4\) and atomic number by \(2\). \[ \beta^{-}\text{-particle} \] emission does not change mass number, but increases atomic number by \(1\).

Step 1: Find the number of \(\alpha\)-particles using mass number.
Initial mass number: \[ 200 \] Final mass number: \[ 168 \] Decrease in mass number: \[ 200-168=32 \] Since one \(\alpha\)-particle decreases mass number by \(4\), \[ \text{Number of } \alpha\text{-particles}=\frac{32}{4}=8 \] So, \[ \alpha = 8 \]

Step 2: Find atomic number after emitting \(8\alpha\)-particles.
Initial atomic number: \[ 90 \] One \(\alpha\)-particle decreases atomic number by \(2\). So, \(8\alpha\)-particles decrease atomic number by: \[ 8 \times 2=16 \] Therefore, atomic number after \(\alpha\)-emission: \[ 90-16=74 \]

Step 3: Find the number of \(\beta\)-particles.
Final atomic number is: \[ 80 \] After \(\alpha\)-emission, atomic number is: \[ 74 \] Now it must increase from \(74\) to \(80\). Increase required: \[ 80-74=6 \] Since one \(\beta^{-}\)-particle increases atomic number by \(1\), \[ \beta = 6 \]

Step 4: Final conclusion.
Hence, the respective numbers of \(\alpha\) and \(\beta\) particles are: \[ \boxed{8\text{ and }6} \]