A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If $M(A, Z)$ denotes the mass of a single neutral atom of an element with mass number $A$ and atomic number $Z$, then the minimal condition that the $\beta$ decay $X_Z^A \rightarrow Y_{Z+1}^A+\beta^{-}+\bar{v}_e$ will occur is ( $m_e$ denotes the mass of the $\beta$ particle and the neutrino mass $m_v$ can be neglected)
$M(A, Z) > M(A, Z+1)+m_e$
$M(A, Z) > M(A, Z+1)$
$M(A, Z) > M(A, Z+1)+Z m_e$
$M(A, Z) > M(A, Z+1)-m_e$
How many alpha and beta particles are emitted when Uranium ${ }_{92} U ^{238}$ decays to lead ${ }_{82} Pb ^{206}$ ?
${U^{238}}$ decays into $T{h^{234}}$ by the emission of an $\alpha - $ particle. There follows a chain of further radioactive decays, either by $\alpha - $ decay or by $\beta $ - decay. Eventually a stable nuclide is reached and after that, no further radioactive decay is possible. Which of the following stable nuclides is the and product of the ${U^{238}}$ radioactive decay chain
Match List $I$ of the nuclear processes with List $II$ containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists :
List $I$ | List $II$ |
$P.$ $\quad$ Alpha decay | $1.$ $\quad{ }_8^{15} 0 \rightarrow{ }_7^{15} N +\ldots \ldots$. |
$Q.$ $\quad$ $\beta^{+}$decay | $2.$ $\quad{ }_{92}^{238} U \rightarrow{ }_{90}^{234} Th +\ldots \ldots$. |
$R.$ $\quad$Fission | $3.$ $\quad{ }_{83}^{185} Bi \rightarrow{ }_{82}^{184} Pb +\ldots \ldots$. |
$S.$ $\quad$Proton emission | $4.$ $\quad{ }_{94}^{239} Pu \rightarrow{ }_{57}^{140} La +\ldots \ldots$. |
Codes: $ \quad \quad P \quad Q \quad R \quad S $
Before the neutrino hypothesis, the beta decay process was thought to be the transition, $n \to p + {e^ - }$ If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.
The mass of a nucleus ${ }_Z^A X$ is less that the sum of the masses of $(A-Z)$ number of neutrons and $Z$ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass $M$ can break into two light nuclei of masses $m_1$ and $m_2$ only if $\left(m_1+m_2\right)M^{\prime}$. The masses of some neutral atoms are given in the table below:
${ }_1^1 H$ | $1.007825 u$ | ${ }_2^1 H$ | $2.014102 u$ | ${ }_3^1 H$ | $3.016050 u$ | ${ }_2^4 He$ | $4.002603 u$ |
${ }_3^6 Li$ | $6.015123 u$ | ${ }_7^3 Li$ | $7.016004 u$ | ${ }_70^30 Zn$ | $69.925325 u$ | ${ }_{34}^{82} Se$ | $81.916709 u$ |
${ }_{64}^{152} Gd$ | $151.919803 u$ | ${ }_{206}^{82} Gd$ | $205.974455 u$ | ${ }_{209}^{83} Bi$ | $208.980388 u$ | ${ }_{84}^{210} Po$ | $209.982876 u$ |
$1.$ The correct statement is:
$(A)$ The nucleus ${ }_3^6 Li$ can emit an alpha particle
$(B)$ The nucleus ${ }_{84}^{210} P_0$ can emit a proton
$(C)$ Deuteron and alpha particle can undergo complete fusion.
$(D)$ The nuclei ${ }_{30}^{70} Zn$ and ${ }_{34}^{82} Se$ can undergo complete fusion.
$2.$ The kinetic energy (in $keV$ ) of the alpha particle, when the nucleus ${ }_{84}^{210} P _0$ at rest undergoes alpha decay, is:
$(A)$ $5319$ $(B)$ $5422$ $(C)$ $5707$ $(D)$ $5818$
Give the answer question $1$ and $2.$