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: Effect of Alpha ( \( \alpha \) ) Decay

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

When element \( X \) undergoes alpha decay:

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

Thus, the new nucleus \( Y \) has:

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

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

Step 2: Effect of Positron (\( e^+ \)) Emission

Positron emission (\( \beta^+ \)) decreases the atomic number by 1 without changing the mass number:

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

Thus, \( Z \) has:

Mass number = 286 (unchanged)

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

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

Beta-minus (\( \beta^- \)) emission increases the atomic number by 1 without affecting the mass number:

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

Thus, \( P \) has:

Mass number = 286 (unchanged)

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

Step 4: Effect of Electron Capture (\( e^- \))

Electron capture decreases the atomic number by 1, keeping the mass number unchanged:

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

Thus, \( Q \) has:

Mass number = 286 (unchanged)

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

Conclusion

The final product \( Q \) has mass number 286 and atomic number 81, which corresponds to option (4).