If $_{92}{U^{238}}$ undergoes successively $8 \alpha$- decays and $6 \beta$- decays, then resulting nucleus is
$_{82}{U^{206}}$
$_{82}P{b^{206}}$
$_{82}{U^{210}}$
$_{82}{U^{214}}$
A radioactive nucleus undergoes a series of decay according to the scheme
$A\xrightarrow{\alpha }{{A}_{1}}\xrightarrow{\beta }{{A}_{2}}\xrightarrow{\alpha }{{A}_{3}}\xrightarrow{\gamma }{{A}_{4}}$
If the mass number and atomic number of $A$ are $180$ and $72$ respectively, then what are these number for $A_4$
In the given nuclear reaction, the element $X$ is :
${ }_{11}^{22} Na \rightarrow X + e ^{+}+v$
What is gamma decay ? Explain by proper example.
Three $\alpha - $ particles and one $\beta - $ particle decaying takes place in series from an isotope $_{88}R{a^{238}}$. Finally the isotope obtained will be
${ }_{92}^{235} U$ atom disintegrates to ${ }_{82}^{207} Pb$ with a half-life of $10^9 yr$. In the process, it emits $7 \alpha$ particles and $n \beta^{-}$particles. Here, $n$ is