$A$ nucleus with mass number $220$ initially at rest emits an $\alpha -$ particle. If the $Q$ value of the reaction is $7.8 \, MeV$ and subsequently after the emission of the $\alpha -$ particle,a photon of energy $1.2 \, MeV$ is also emitted. The kinetic energy of the $\alpha -$ particle is ? .......... $MeV$

In a nuclear fusion process,the mass defect is $0.05\%$. How much energy will be released from $1 \, kg$ of mass?

$A$ fission reaction is given by ${ }_{92}^{236} U \rightarrow{ }_{54}^{140} Xe +{ }_{38}^{94} Sr + x + y$,where $x$ and $y$ are two particles. Considering ${ }_{92}^{236} U$ to be at rest,the kinetic energies of the products are denoted by $K_{Xe}, K_{Sr}, K_x (2 \ MeV)$ and $K_y (2 \ MeV)$,respectively. Let the binding energies per nucleon of ${ }_{92}^{236} U, { }_{54}^{140} Xe$ and ${ }_{38}^{94} Sr$ be $7.5 \ MeV, 8.5 \ MeV$ and $8.5 \ MeV$ respectively. Considering different conservation laws,the correct option$(s)$ is(are):

Complete the following nuclear reaction:
${ }_{0}^{1} n+{ }_{92}^{235} U \rightarrow{ }_{92}^{236} U \rightarrow$ $\qquad$ $+{ }_{41}^{99} Nb+$ $\qquad$.

The rest mass of an electron as well as that of a positron is $0.51 \, MeV$. When an electron and a positron annihilate, they produce gamma-rays of wavelength(s):

In a nuclear fission reaction: