The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ${ }_{7}^{16} N +{ }_{2}^{4} He \rightarrow{ }_{1}^{1} H +{ }_{8}^{19} O$ in a laboratory frame is $n$ (in $MeV$). Assume that ${ }_{7}^{16} N$ is at rest in the laboratory frame. The masses of ${ }_{7}^{16} N, { }_{2}^{4} He, { }_{1}^{1} H$ and ${ }_{8}^{19} O$ can be taken to be $16.006 \ u, 4.003 \ u, 1.008 \ u$ and $19.003 \ u$,respectively,where $1 \ u = 930 \ MeV/c^2$. The value of $n$ is. . . . .

The energy released per fission of uranium-$235$ is about $200 \, MeV$. $A$ reactor using $U-235$ as fuel is producing $1000 \, kW$ power. The number of $U-235$ nuclei undergoing fission per second is approximately:

Which of the following is used as a moderator in nuclear reactors?

If $200 \, MeV$ energy is released in the fission of a single nucleus of ${}_{92}^{235}U$,the number of fissions required per second to produce a power of $1 \, kW$ is:

$A$ 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 ${ }^{12} C = 12 \ u$,$1 \ u = 1.66 \times 10^{-27} \ kg$,$c = 3 \times 10^8 \ m/s$]

$A$ certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in the fusion reaction is $0.02866 \, u$. The energy liberated per $u$ is ........... $MeV$. (Given $1 \, u = 931 \, MeV$)