In the options given below, let $E$ denote the rest mass energy of a nucleus and $n$ a neutron. The correct option is:

If the energy released in the fission of one nucleus is $200 \, MeV$, then the number of nuclei required per second in a power plant of $16 \, kW$ will be:

When two nuclei of mass $X$ and $Y$ respectively fuse to form a nucleus of mass $m$ along with the liberation of some energy, then

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

$A$ nuclear reactor provides a power of $300 \, MW$. If the fission of one uranium nucleus ${U^{235}}$ releases $170 \, MeV$ of energy,how many nuclei undergo fission in $1 \, h$?

Consider the fission of $_{92}^{238} U$ by fast neutrons. In one fission event,no neutrons are emitted and the final end products,after the beta decay of the primary fragments,are $_{58}^{140} Ce$ and $_{44}^{99} Ru$. Calculate the $Q$-value for this fission process. The relevant atomic and particle masses are:
$m(_{92}^{238} U) = 238.05079 \; u$
$m(_{58}^{140} Ce) = 139.90543 \; u$
$m(_{44}^{99} Ru) = 98.90594 \; u$
$m(_{0}^{1} n) = 1.008665 \; u$