The energy released in an atom bomb is obtained from?

During the fission of $_{92}U^{235}$,$0.1\%$ of its mass is converted into energy. How much energy is produced from $1 \, kg$ of $_{92}U^{235}$?

Suppose a ${ }_{88}^{226} Ra$ nucleus at rest and in the ground state undergoes $\alpha$-decay to a ${ }_{86}^{222} Rn$ nucleus in its excited state. The kinetic energy of the emitted $\alpha$ particle is found to be $4.44 \text{ MeV}$. The ${ }_{86}^{222} Rn$ nucleus then goes to its ground state by $\gamma$-decay. The energy of the emitted $\gamma$-photon is . . . . . . . $\text{keV}$.
[Given: atomic mass of ${ }_{88}^{226} Ra = 226.005 \text{ u}$,atomic mass of ${ }_{86}^{222} Rn = 222.000 \text{ u}$,atomic mass of $\alpha$ particle $= 4.000 \text{ u}$,$1 \text{ u} = 931 \text{ MeV}/c^2$]

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 $c^2 = 931 \text{ MeV/u}$.

Which reaction is not part of the proton-proton cycle?

According to classical physics,$10^{-15} \ m$ is the distance of closest approach $(d_c)$ for fusion to occur between two protons. $A$ more accurate quantum approach states that $d_c = \frac{\lambda_p}{\sqrt{2}}$,where $\lambda_p$ is the de Broglie wavelength of a proton when they were far apart. Using this quantum approach,find the equation for the temperature $(T_c)$ at the center of a star. [Given: $M_p$ is the mass of the proton,$k$ is the Boltzmann constant,$e$ is the elementary charge]