In the following nuclear rection, $D \stackrel{\alpha}{\longrightarrow} D _{1} \stackrel{\beta-}{\longrightarrow} D _{2} \stackrel{\alpha}{\longrightarrow} D _{3} \stackrel{\gamma}{\longrightarrow} D _{4}$ Mass number of $D$ is $182$ and atomic number is $74$ . Mass number and atomic number of $D _{4}$ respectively will be
$174$ and $71$
$174$ and $69$
$172$ and $69$
$172$ and $71$
The electron emitted in beta radiation originates from
If $_{92}{U^{238}}$ undergoes successively $8 \alpha$- decays and $6 \beta$- decays, then resulting nucleus is
$\gamma$-decay occurs when
Consider a $\beta$ decay reaction
${}_1^3H \to {}_2^3He + {e^{ - 1}} + \bar v$
Atomic mass of ${}_1^3H$and ${}_2^3He$ are $3.016050\,u$ and $3.016030\,u$. Find the maximum possible energy of electron ....... $MeV$
The isotope ${ }_5^{12} \mathrm{~B}$ having a mass $12.014 \mathrm{u}$ undergoes $\beta$-decay to ${ }_6^{12} \mathrm{C} .{ }_6^{12 .}$ has an excited state of the nucleus $\left({ }_6^{12} \mathrm{C}^*\right)$ at $4.041 \mathrm{MeV}$ above its ground state. If ${ }_5^{12} \mathrm{~F}$ decays to ${ }_6^{12} \mathrm{C}^*$, the maximum kinetic energy of the $\beta$-particle in units of $\mathrm{MeV}$ is ( $1 \mathrm{u}=931.5 \mathrm{MeV} / c^2$, where $c$ is the speed of light in vacuum).