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Expert-verified Found in: Page 1332 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A star converts all its hydrogen to helium, achieving a 100% helium composition. Next, it converts the helium to carbon via the triple-alpha process, ${}^{{\mathbf{4}}}{\mathbit{H}}{\mathbit{e}}{\mathbf{+}}{}^{{\mathbf{4}}}{\mathbit{H}}{\mathbit{e}}{\mathbf{+}}{}^{{\mathbf{4}}}{\mathbit{H}}{\mathbit{e}}{{\mathbf{\to }}}^{{\mathbf{12}}}{\mathbit{C}}{\mathbf{+}}{\mathbf{7}}{\mathbf{.}}{\mathbf{27}}{\mathbf{}}{\mathbit{M}}{\mathbit{e}}{\mathbit{V}}$The mass of the star is role="math" localid="1661754478822" ${\mathbf{4}}{\mathbf{.}}{\mathbf{6}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{32}}}{\mathbit{k}}{\mathbit{g}}$, and it generates energy at the rate of ${\mathbf{5}}{\mathbf{.}}{\mathbf{3}}{\mathbit{x}}{{\mathbf{10}}}^{{\mathbf{3}}}{\mathbf{}}{\mathbit{W}}$. How long will it take to convert all the helium to carbon at this rate?

The required time is $1.6×{10}^{8}yr$.

See the step by step solution

## Step 1: Describe the expression for the time

Let m be the mass of helium, the number of three-helium atoms in this mass equals the number of moles multiplied by the number of atoms in one mole, where the number of moles equals the mass divided by three the molar mass of the helium.

${{\mathbit{N}}}_{\mathbf{3}\mathbf{H}\mathbf{e}}{\mathbf{=}}\frac{{\mathbf{m}}_{\mathbf{H}\mathbf{e}}{\mathbf{N}}_{\mathbf{A}}}{\mathbf{3}{\mathbf{M}}_{\mathbf{H}\mathbf{e}}}$

Let be the energy release per fusion, then the total energy released by fusion equals the number of fusions multiplied by ${{\mathbit{N}}}_{\mathbf{3}\mathbf{He}}$.

${{\mathbit{E}}}_{\mathbf{f}\mathbf{u}\mathbf{s}\mathbf{i}\mathbf{o}\mathbf{n}}{\mathbf{=}}{{\mathbit{N}}}_{\mathbf{3}\mathbf{H}\mathbf{e}}{\mathbit{Q}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{=}}\frac{{\mathbf{m}}_{\mathbf{H}\mathbf{e}}{\mathbf{N}}_{\mathbf{A}}\mathbf{Q}}{\mathbf{3}{\mathbf{M}}_{\mathbf{H}\mathbf{e}}}$

The energy also can be written as follows:

${{\mathbit{E}}}_{\mathbf{f}\mathbf{u}\mathbf{s}\mathbf{i}\mathbf{o}\mathbf{n}}{\mathbf{=}}{\mathbit{P}}{\mathbit{t}}\phantom{\rule{0ex}{0ex}}\frac{{\mathbf{m}}_{\mathbf{H}\mathbf{e}}{\mathbf{N}}_{\mathbf{A}}\mathbf{Q}}{\mathbf{3}{\mathbf{M}}_{\mathbf{H}\mathbf{e}}}{\mathbf{=}}{\mathbit{P}}{\mathbit{t}}\phantom{\rule{0ex}{0ex}}{\mathbit{t}}{\mathbf{=}}\frac{{\mathbf{m}}_{\mathbf{H}\mathbf{e}}{\mathbf{N}}_{\mathbf{A}}\mathbf{Q}}{\mathbf{3}{\mathbf{M}}_{\mathbf{H}\mathbf{e}}\mathbf{P}}$

## Step 2: Find the time required to convert all the helium to carbon

Substitute all the known values in equation (1).

$t=\frac{\left(4.6×{10}^{35}g\right)\left(6.022×{10}^{23}mo{l}^{-1}\right)\left(7.27MeV\right)\left(1.602×{10}^{-13}J/MeV\right)}{3\left(4.0g/mol\right)\left(5.3×{10}^{30}W\right)}\phantom{\rule{0ex}{0ex}}=5.07×{10}^{15}s\phantom{\rule{0ex}{0ex}}=\frac{5.07×{10}^{15}s}{3.154×{10}^{7}s/yr}\phantom{\rule{0ex}{0ex}}=1.6×{10}^{8}yr$

Therefore, the required time is $1.6×{10}^{8}yr$. ### Want to see more solutions like these? 