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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 # Verify the three Q values reported for the reactions given in Fig. 43-11.The needed atomic and particle masses are$\begin{array}{llll}{}^{\mathbf{1}}\mathbf{H}\mathbf{e}& \mathbf{1}\mathbf{.}\mathbf{007825}\mathbf{u}& {}^{\mathbf{4}}\mathbf{H}\mathbf{e}& \mathbf{4}\mathbf{.}\mathbf{002603}\mathbf{u}\\ {}^{\mathbf{2}}\mathbf{H}& \mathbf{2}\mathbf{.}\mathbf{014102}\mathbf{u}& {\mathbf{e}}^{\mathbf{±}}& \mathbf{0}\mathbf{.}\mathbf{0005486}\mathbf{u}\end{array}\phantom{\rule{0ex}{0ex}}{}^{{\mathbf{3}}}{\mathbit{H}}{\mathbit{e}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{3}}{\mathbf{.}}{\mathbf{016029}}{\mathbit{u}}$(Hint: Distinguish carefully between atomic and nuclear masses, and take the positrons properly into account.) All the energy values are verified.

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

## Step 2: Verify the three Q values reported for the reactions

Consider the first reaction.

${}^{1}H+{}^{1}H{\to }^{2}H+{e}^{+}+v$

The energy equation can be written as follows.

${Q}_{1}=\left(2{m}_{p}-{m}_{d}-{m}_{{e}^{+}}\right){c}^{2}$

Here,${m}_{p}$ is the mass of the proton,${m}_{d}$ is the mass of deuteron, and${m}_{{e}^{+}}$ is the mass of positron.

Rewrite the equation as follows.

role="math" localid="1661923054546" ${Q}_{1}=\left[2\left({m}_{1H}-{m}_{{e}^{+}}\right)-\left({m}_{2H}-{m}_{{e}^{+}}\right)-{m}_{{e}^{+}}\right]{c}^{2}\phantom{\rule{0ex}{0ex}}=\left[2{m}_{1H}-{m}_{2H}-2{m}_{{e}^{+}}\right]{c}^{2}......\left(1\right)$

Substitute all the known values in equation (1).

${Q}_{1=}\left[2\left(1.007825u\right)-2.014102u-2\left(0.0005486u\right)\right]\left(931.5MeV/u\right)\phantom{\rule{0ex}{0ex}}=0.41984568MeV\phantom{\rule{0ex}{0ex}}\approx 0.42MeV$

Therefore, the energy released is $0.42MeV$.

Consider the second reaction.

${}^{2}H+{}^{1}H{\to }^{3}He+y$

The energy equation can be written as follows.

${Q}_{2}=\left({m}_{2H}+{m}_{1H}-{m}_{3He}\right){c}^{2}.....\left(2\right)$

Substitute all the known values in equation (2).

${Q}_{2}=\left[\left(2.014102u\right)+\left(1.007825u\right)-\left(3.016029u\right)\right]\left(931.5MeV/u\right)\phantom{\rule{0ex}{0ex}}=5.493987MeV\phantom{\rule{0ex}{0ex}}\approx 5.49MeV$

Therefore, the energy released is$5.49MeV$ .

Consider the third reaction.

${}^{3}He+{}^{3}He{\to }^{4}He+={+}^{1}H{+}^{1}H$

The energy equation can be written as follows.

${Q}_{3}=\left[2\left({m}_{3He}\right)-{m}_{4He}-2\left({m}_{1H}\right)\right]{c}^{2}......\left(3\right)$

Substitute all the known values in equation (3).

${Q}_{3}=\left[2\left(3.016029u\right)-\left(4.002603u\right)-\left(1.007825u\right)\right]\left(931.5MeV/u\right)\phantom{\rule{0ex}{0ex}}=12.8593575MeV\phantom{\rule{0ex}{0ex}}\approx 12.86MeV$

Therefore, the energy released is $12.86MeV$.

Therefore, all the energy values are verified. ### Want to see more solutions like these? 