$A$ star has $10^{40}$ deuterons. It produces energy via the processes:
$_1H^2 + _1H^2 \to _1H^3 + p$
$_1H^2 + _1H^3 \to _2He^4 + n$
If the average power radiated by the star is $10^{16} \ W$, the deuteron supply of the star is exhausted in a time of the order of:
Given:
Mass of $_1H^2 = 2.014 \ amu$
Mass of $_2He^4 = 4.001 \ amu$
Mass of proton = $1.007 \ amu$
Mass of neutron = $1.008 \ amu$
$_1H^2 + _1H^2 \to _1H^3 + p$
$_1H^2 + _1H^3 \to _2He^4 + n$
If the average power radiated by the star is $10^{16} \ W$, the deuteron supply of the star is exhausted in a time of the order of:
Given:
Mass of $_1H^2 = 2.014 \ amu$
Mass of $_2He^4 = 4.001 \ amu$
Mass of proton = $1.007 \ amu$
Mass of neutron = $1.008 \ amu$
- A$10^6 \ s$
- B$10^8 \ s$
- C$10^{12} \ s$
- D$10^{16} \ s$
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Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below:
${ }_{1}^{2} X +{ }_{1}^{2} X ={ }_{2}^{4} Y$
The binding energies per nucleon of ${ }_{1}^{2} X$ and ${ }_{2}^{4} Y$ are $1.1 \, MeV$ and $7.6 \, MeV$ respectively. The energy released in this process is in $MeV$.
${ }_{1}^{2} X +{ }_{1}^{2} X ={ }_{2}^{4} Y$
The binding energies per nucleon of ${ }_{1}^{2} X$ and ${ }_{2}^{4} Y$ are $1.1 \, MeV$ and $7.6 \, MeV$ respectively. The energy released in this process is in $MeV$.
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View SolutionThe 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).
The binding energy per nucleon for deuteron $(_1H^2)$ and helium $(_2He^4)$ are $1.1 \, MeV$ and $7 \, MeV$ respectively. When two deuterons fuse to form a helium nucleus,the energy released is ........... $MeV$.
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