What is the energy released in the process $3 \, _{2}^{4}\text{He} \to _{6}^{12}\text{C}$ (in $, \text{MeV}$)? (Mass of $_{2}^{4}\text{He} = 4.002604 \, \text{amu}$,mass of $_{6}^{12}\text{C} = 12 \, \text{amu}$)

Energy in the sun is generated mainly by

Nuclear fission was discovered by

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

Assertion : Energy is released in nuclear fission.
Reason : Total binding energy of the fission fragments is larger than the total binding energy of the parent nucleus.

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