Question 353: Atoms & Nuclei

Question

A star has $$100 \%$$ helium composition. It starts to convert three $${ }^4 \mathrm{He}$$ into one $${ }^{12} \mathrm{C}$$ via triple alpha process as $${ }^4 \mathrm{He}+{ }^4 \mathrm{He}+{ }^4 \mathrm{He} ightarrow{ }^{12} \mathrm{C}+\mathrm{Q}$$. The mass of the star is $$2.0 	imes 10^{32} \mathrm{~kg}$$ and it generates energy at the rate of $$5.808 	imes 10^{30} \mathrm{~W}$$. The rate of converting these $${ }^4 \mathrm{He}$$ to $${ }^{12} \mathrm{C}$$ is $$\mathrm{n} 	imes 10^{42} \mathrm{~s}^{-1}$$, where $$\mathrm{n}$$ is _________. [ Take, mass of $${ }^4 \mathrm{He}=4.0026 \mathrm{u}$$, mass of $${ }^{12} \mathrm{C}=12 \mathrm{u}$$]

Accepted Answer

To determine the rate of converting $${ }^4 \mathrm{He}$$ to $${ }^{12} \mathrm{C}$$, we need to calculate the energy released per reaction and use the given power production of the star to find the rate of reactions. The relevant nuclear reaction is:
$${ }^4 \mathrm{He} + { }^4 \mathrm{He} + { }^4 \mathrm{He} ightarrow { }^{12} \mathrm{C} + \mathrm{Q}$$
The masses involved in the reaction are given:
Mass of $${ }^4 \mathrm{He} = 4.0026 \, \mathrm{u}$$
Mass of $${ }^{12} \mathrm{C} = 12 \, \mathrm{u}$$
First, we calculate the mass defect (difference between the mass of reactants and products) which will give us the energy released in each reaction:
Mass of reactants: $$3 	imes 4.0026 \, \mathrm{u} = 12.0078 \, \mathrm{u}$$
Mass of product: $$12 \, \mathrm{u}$$
Mass defect: $$12.0078 \, \mathrm{u} - 12 \, \mathrm{u} = 0.0078 \, \mathrm{u}$$
We use Einstein's mass-energy equivalence principle, $$E = mc^2$$, to find the energy released per reaction. The conversion factor between atomic mass units and energy is $$1 \, \mathrm{u} = 931.5 \, \mathrm{MeV}$$.
Energy released per reaction: $$0.0078 \, \mathrm{u} 	imes 931.5 \, \mathrm{MeV/u} = 7.2627 \, \mathrm{MeV}$$
We convert this energy into joules. $$1 \, \mathrm{MeV} = 1.60218 	imes 10^{-13} \, \mathrm{J}$$:
Energy per reaction: $$7.2627 \, \mathrm{MeV} 	imes 1.60218 	imes 10^{-13} \, \mathrm{J/MeV} = 1.163 	imes 10^{-12} \, \mathrm{J}$$
The power generated by the star is given as $$5.808 	imes 10^{30} \, \mathrm{W}$$. The rate of the reaction is the power divided by the energy per reaction:
$$	ext{Rate of reactions} = \frac{	ext{Power}}{	ext{Energy per reaction}}$$
$$ 	ext{Rate} = \frac{5.808 	imes 10^{30} \, \mathrm{W}}{1.163 	imes 10^{-12} \, \mathrm{J}}$$
$$ 	ext{Rate} = 4.99 	imes 10^{42} \, \mathrm{s^{-1}} \simeq 5 	imes 10^{42} \mathrm{~s}^{-1}$$
Thus, the rate of converting $${ }^4 \mathrm{He}$$ to $${ }^{12} \mathrm{C}$$ is:
$$ \mathrm{n} = 5$$

Atoms & Nuclei

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