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).