For the nuclear fusion reaction ${ }_1^2 H +{ }_1^3 H \rightarrow{ }_2^4 He +{ }_0^1 n$,the temperature to which the gases must be heated is $3.7 \times 10^9 \, K$. The potential energy between two nuclei is closest to ........ $J$ (Boltzmann's constant $k = 1.38 \times 10^{-23} \, J/K$).

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:

$A$ nucleus with mass number $220$ initially at rest emits an alpha particle. If the $Q$ value of the reaction is $5.5 \ MeV$,calculate the kinetic energy of the alpha particle. (in $MeV$)

The isotope ${}_{5}^{12}B$ having a mass $12.014 \text{ u}$ undergoes $\beta$-decay to ${}_{6}^{12}C$. ${}_{6}^{12}C$ has an excited state of the nucleus $({}_{6}^{12}C^*)$ at $4.041 \text{ MeV}$ above its ground state. If ${}_{5}^{12}B$ decays to ${}_{6}^{12}C^*$, the maximum kinetic energy of the $\beta$-particle in units of $\text{MeV}$ is ($1 \text{ u} = 931.5 \text{ MeV}/c^2$, where $c$ is the speed of light in vacuum).

If $200 \text{ MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{236} U$,the number of nuclei that must undergo fission to release an energy of $1000 \text{ J}$ 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 ${ }^{12} C = 12 \ u$,$1 \ u = 1.66 \times 10^{-27} \ kg$,$c = 3 \times 10^8 \ m/s$]