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Q46P

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Found in: Page 1333

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# In certain stars the carbon cycle is more effective than the proton–proton cycle in generating energy.This carbon cycle is${}^{{\mathbf{12}}}{\mathbf{C}}{\mathbf{+}}{}^{{\mathbf{1}}}{\mathbf{H}}{{\mathbf{\to }}}^{{\mathbf{13}}}{\mathbf{N}}{\mathbf{+}}{\mathbf{\gamma }}{\mathbf{,}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{{\mathbf{Q}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{95}}{\mathbf{}}{\mathbf{MeV}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{}^{{\mathbf{13}}}{\mathbf{N}}{{\mathbf{\to }}}^{{\mathbf{13}}}{\mathbf{C}}{\mathbf{+}}{{\mathbf{e}}}^{{\mathbf{+}}}{\mathbf{+}}{\mathbf{v}}{\mathbf{,}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{{\mathbf{Q}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{19}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{}^{{\mathbf{13}}}{\mathbf{C}}{\mathbf{+}}{}^{{\mathbf{1}}}{\mathbf{H}}{\mathbf{\to }}^{\mathbf{14}}{\mathbf{N}}{\mathbf{+}}{\mathbf{\gamma }}{\mathbf{,}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{{\mathbf{Q}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{7}}{\mathbf{.}}{\mathbf{55}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{}^{{\mathbf{14}}}{\mathbf{C}}{\mathbf{+}}{}^{{\mathbf{1}}}{\mathbf{H}}{\mathbf{\to }}^{\mathbf{15}}{\mathbf{O}}{\mathbf{+}}{\mathbf{\gamma }}{\mathbf{,}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{{\mathbf{Q}}}_{{\mathbf{4}}}{\mathbf{=}}{\mathbf{7}}{\mathbf{.}}{\mathbf{30}}{\mathbf{,}}{\phantom{\rule{0ex}{0ex}}}^{{\mathbf{15}}}{\mathbf{O}}{\mathbf{}}{{\mathbf{\to }}}^{{\mathbf{15}}}{\mathbf{N}}{\mathbf{+}}{{\mathbf{e}}}^{{\mathbf{+}}}{\mathbf{+}}{\mathbf{v}}{\mathbf{,}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{{\mathbf{Q}}}_{{\mathbf{5}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{73}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{}^{{\mathbf{15}}}{\mathbf{C}}{\mathbf{+}}{}^{{\mathbf{1}}}{\mathbf{H}}{\mathbf{\to }}^{\mathbf{12}}{\mathbf{C}}{{\mathbf{+}}}^{{\mathbf{4}}}{\mathbf{He}}{\mathbf{,}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{{\mathbf{Q}}}_{{\mathbf{6}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{97}}$(a) Show that this cycle is exactly equivalent in its overall effects to the proton–proton cycle of Fig. 43-11. (b) Verify that the two cycles, as expected, have the same Q value.

(a) The carbon cycle is exactly equivalent to proton-proton cycle.

(b) It is verified that the two cycles have the same Q value.

See the step by step solution

## Step 1: Describe the expression for energy

The expression for energy is given by,

${\mathbit{Q}}{\mathbf{=}}{\mathbf{-}}{\mathbf{∆}}{\mathbit{m}}{{\mathbit{c}}}^{{\mathbf{2}}}$

Here, is the energy release in a reaction, ${\mathbf{∆}}{\mathbit{m}}$ is the mass difference between the parent nuclei and the daughter nuclei, and c is the velocity of light.

## Step 2: Show that carbon-carbon cycle is exactly equivalent to the proton–proton cycle(a)

If can be carefully observed thatin the carbon-carbon cycle, the products are two positrons $\left(2{e}^{+}\right)$ , two neutrinos $\left(2v\right)$ , and one helium nucleus $\left({}^{4}He\right)$ .

The same products are also obtained in the proton-proton cycle.

Therefore, the carbon cycle is exactly equivalent to proton-proton cycle.

## Step 3: Verify that the two cycles have the same Q value(b)

The value of the energy is the sum of the Q values of the carbon cycle.

${Q}_{c}=\sum _{n=1}^{6}{Q}_{n}\phantom{\rule{0ex}{0ex}}={Q}_{1}+{Q}_{2}+{Q}_{3}+{Q}_{4}+{Q}_{5}+{Q}_{6}\phantom{\rule{0ex}{0ex}}=1.95Mev+1.19Mev+7.55Mev+7.30Mev+1/73Mev+4.97Mev\phantom{\rule{0ex}{0ex}}=24.7Mev$

The Q value is same for both cycles.

Therefore, it is verified that the two cycles have the same Q value.