During a negative beta decay
An atomic electron is ejected
An electron which is already present within the nucleus is ejected
A neutron in the nucleus decays emitting an electron
A part of the binding energy is converted into electron
What is the respective number of $\alpha $ and $\beta $ particles emitted in the following radioactive decay
$_{90}{X^{200}}{ \to _{80}}{Y^{168}}$
A nucleus of lead $Pb _{82}^{214}$ emits two electrons followed by an $\alpha$-particle. The resulting nucleus will have
In the nuclear reaction $_{85}{X^{297}} \to Y + 4\alpha ,\;Y$ is
${ }_{82}^{290} X \xrightarrow{\alpha} Y \xrightarrow{e^{+}} Z \xrightarrow{\beta^{-}} P \xrightarrow{e^{-}} Q$
In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are
In pair annihilation, an electron and a positron destroy each other to produce gamma radiation. How is the momentum conserved ?