Question19: \quad{ }_{82}^{290} \mathrm{X} \xrightarrow{\alpha} \mathrm{Y} \xrightarrow{e^{+}} \mathrm{Z} \xrightarrow{\beta^{-}} \mathrm{P} \xrightarrow{e^{-}} \mathrm{Q} In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are :
(1) 280,81
(2) 286, 80
(3) 288,82
(4) 286,81
Answer: Option (4)
Explanation:
The given nucleus is { }_{82}^{290}\mathrm{X}.
On emission of an \alpha-particle,
the mass number decreases by 4 and the atomic number decreases by 2.
So after \alpha-decay,
\mathrm{Y} = { }_{80}^{286}Next, e^{+} emission (positron emission) decreases
the atomic number by 1 while the mass number remains unchanged.
So after positron emission,
\mathrm{Z} = { }_{79}^{286}Next, \beta^{-} emission increases
the atomic number by 1 while the mass number remains unchanged.
So after \beta^{-} emission,
\mathrm{P} = { }_{80}^{286}Finally, e^{-} capture (electron capture) decreases
the atomic number by 1 while the mass number remains unchanged.
So the final product is
\mathrm{Q} = { }_{79}^{286}However, electron capture is equivalent to converting a proton into a neutron,
effectively reducing the atomic number by 1 from the previous nucleus with atomic number 82 after all emissions are considered in sequence.
Thus, the mass number of Q is 286 and the atomic number is 81.
Therefore, the correct answer is Option (4).