An artificial radioactive decay series begins with unstable $_{94}^{241}Pu$. The stable nuclide obtained after eight $\alpha - $ decays and five $\beta - $ decays is

$\alpha$ -particle consists of 

Atomic mass number of an element thorium is $232$ and its atomic number is $90$. The end product of this radioactive element is an isotope of lead (atomic mass $208$ and atomic number $82$). The number of alpha and beta particles emitted is

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:

$1.$ 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 $2.$

A radioactive element $X$ emits six $\alpha$-particles and four $\beta$-particles leading to ond product ${ }_{82}^{208} Pb$. $X$ is

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,