In a material medium,when a positron meets an electron,both particles annihilate,leading to the emission of two gamma-ray photons. This process forms the basis of an important diagnostic procedure called:

When $_3^7Li$ nuclei are bombarded by protons, and the resultant nuclei are $_4^8Be$, the emitted particles will be

On bombarding $U^{235}$ by slow neutrons,$200 \text{ MeV}$ of energy is released. If the power output of an atomic reactor is $1.6 \text{ MW}$,then the rate of fission will be:

The source of the Sun's energy is

The fission of $_{92}^{235}U$ can be triggered by the absorption of a slow neutron by a nucleus. Similarly,a slow proton can also be used. This statement is

The disintegration energy $Q$ for the nuclear fission of ${ }^{235} U \rightarrow{ }^{140} Ce+{ }^{94} Zr+n$ is $\_ \text{MeV}$.
Given atomic masses of:
${ }^{235} U: 235.0439 \text{ u}, { }^{140} Ce: 139.9054 \text{ u},$
${ }^{94} Zr: 93.9063 \text{ u}, n: 1.0086 \text{ u},$
Value of $c^2 = 931 \text{ MeV/u}$.