A nucleus of lead $Pb _{82}^{214}$ emits two electrons followed by an $\alpha$-particle. The resulting nucleus will have
$82$ protons and $128$ neutrons
$80$ protons and $130$ neutrons
$82$ protons and $130$ neutrons
$78$ protons and $134$ neutrons
Consider the following nuclear reactions:
$I$. ${ }_{7}^{14} N +{ }_{2}^{4} He \longrightarrow{ }_{8}^{17} O + X$
$II$. ${ }_{4}^{9} Be +{ }_{2}^{4} H \longrightarrow{ }_{6}^{12} He +Y$
Then,
An element $A$ decays into element $C$ by a two step process :
$A \to B + {\;_2}H{e^4}$
$B \to C + \;2{e^ - }$
Then
During a negative beta decay
Beta rays emitted by a radioactive material are
$\alpha$ -particle consists of