{ }_{82}^{290} X xrightarrow{alpha} Y xrighta | vedclass

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(C) $1$. Alpha decay ($\alpha$): Mass number decreases by $4$, atomic number decreases by $2$.${ }_{82}^{290} X \xrightarrow{\alpha} { }_{80}^{286} Y$

Vedclass · Accepted Answer

$286, 81$

Vedclass · Answer

(C) $1$. Alpha decay ($\alpha$): Mass number decreases by $4$, atomic number decreases by $2$.${ }_{82}^{290} X \xrightarrow{\alpha} { }_{80}^{286} Y$$2$. Positron emission ($e^{+}$): Mass number remains same, atomic number decreases by $1$.${ }_{80}^{286} Y \xrightarrow{e^{+}} { }_{79}^{286} Z$$3$. Beta decay ($\beta^{-}$): Mass number remains same, atomic number increases by $1$.${ }_{79}^{286} Z \xrightarrow{\beta^{-}} { }_{80}^{286} P$$4$. Electron emission ($e^{-}$): This is equivalent to $\beta^{-}$ decay. Mass number remains same, atomic number increases by $1$.${ }_{80}^{286} P \xrightarrow{e^{-}} { }_{81}^{286} Q$Thus, for the final product $Q$, the mass number $A = 286$ and the atomic number $Z = 81$.

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

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