In the following reaction the value of $‘X’$ is $_7{N^{14}}{ + _2}H{e^4}\, \to \,X{ + _1}{H^1}$
$_8{N^{17}}$
$_8{O^{17}}$
$_7{O^{16}}$
$_7{N^{16}}$
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 $A_4$
An atom of mass number $15$ and atomic number $7$ captures an $\alpha - $ particle and then emits a proton. The mass number and atomic number of the resulting product will respectively be
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
The radioactive decay of uranium into thorium is expressed by the equation $_{92}^{238}U \to _{90}^{234}Th + X,$ where $'X'$ is
In the nuclear reaction: $X(n,\,\alpha ){\,_3}L{i^7}$ the term $X$ will be