What is the approximate power output of a $_{92}U^{235}$ reactor if it consumes $2 \ kg$ of fuel in $30 \ days$ and each fission releases $185 \ MeV$ of usable energy? (Avogadro's number $N_A = 6.02 \times 10^{26} \text{ atoms/kmol}$)

In an $\alpha -$ decay,the kinetic energy of the $\alpha -$ particle is $48 \ MeV$ and the $Q$ value of the reaction is $50 \ MeV$. The mass number of the mother nucleus is (assume that the daughter nucleus is in the ground state).

$A$ star has $100 \%$ helium composition. It starts to convert three ${ }^4 He$ into one ${ }^{12} C$ via the triple alpha process as ${ }^4 He + { }^4 He + { }^4 He \rightarrow { }^{12} C + Q$. The mass of the star is $2.0 \times 10^{32} \ kg$ and it generates energy at the rate of $5.808 \times 10^{30} \ W$. The rate of converting these ${ }^4 He$ nuclei to ${ }^{12} C$ is $n \times 10^{42} \ s^{-1}$,where $n$ is. . . . . . . [Take,mass of ${ }^4 He = 4.0026 \ u$,mass of ${ }^{12} C = 12 \ u$,$1 \ u = 1.66 \times 10^{-27} \ kg$,$c = 3 \times 10^8 \ m/s$]

The most suitable element for nuclear fission is the element with an atomic number near

$200\, MeV$ of energy may be obtained per fission of ${U^{235}}$. $A$ reactor is generating $1000\, kW$ of power. The rate of nuclear fission in the reactor is

The binding energy per nucleon for ${}_1^2H$ and ${}_2^4He$ are $1.1 \; MeV$ and $7.1 \; MeV$ respectively. The energy released in $MeV$ when two ${}_1^2H$ nuclei fuse to form one ${}_2^4He$ nucleus is: