Question

\(^{290}_{82}X\stackrel{\alpha}{\rightarrow}Y\stackrel{e^+}{\rightarrow}Z\stackrel{\beta^-}{\rightarrow}P\stackrel{e^-}{\rightarrow}Q\)
In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are :

• A. 280, 81
• B. 286, 80
• C. 288, 82
• D. 286, 81

Answer 1

The Correct Option is D

Step 1: Alpha ( \( \alpha \) ) Decay Impact

An alpha particle (\( \alpha \)) possesses a mass number of 4 and an atomic number of 2.

Following the alpha decay of element \( X \):

$$ {^{290}_{82}X -> ^{286}_{80}Y + ^{4}_{2}\alpha} $$

The resultant nucleus \( Y \) exhibits:

Mass number = \( 290 - 4 = 286 \)

Atomic number = \( 82 - 2 = 80 \)

Step 2: Positron (\( e^+ \)) Emission Impact

Positron emission (\( \beta^+ \)) reduces the atomic number by 1, maintaining the mass number:

$$ {^{286}_{80}Y -> ^{286}_{79}Z + e^+} $$

Nucleus \( Z \) has the following properties:

Mass number = 286 (constant)

Atomic number = \( 80 - 1 = 79 \)

Step 3: Beta-Minus (\( \beta^- \)) Decay Impact

Beta-minus (\( \beta^- \)) emission elevates the atomic number by 1, without altering the mass number:

$$ {^{286}_{79}Z -> ^{286}_{80}P + \beta^-} $$

Nucleus \( P \) has the following properties:

Mass number = 286 (constant)

Atomic number = \( 79 + 1 = 80 \)

Step 4: Electron Capture (\( e^- \)) Impact

Electron capture reduces the atomic number by 1, while the mass number remains constant:

$$ {^{286}_{80}P + e^- -> ^{286}_{81}Q} $$

Nucleus \( Q \) has the following properties:

Mass number = 286 (constant)

Atomic number = \( 80 + 1 = 81 \)

Conclusion

The final product \( Q \) exhibits a mass number of 286 and an atomic number of 81, aligning with option (4).