Verify the three Q values reported for the reactions given in Fig. 43-11.The needed atomic and particle masses are

$$\begin{gathered} \begin{array}{llll}{}^{\mathbf{1}}\mathbf{H}\mathbf{e}& \mathbf{1}\mathbf{.}\mathbf{007825}\mathbf{u}& {}^{\mathbf{4}}\mathbf{H}\mathbf{e}& \mathbf{4}\mathbf{.}\mathbf{002603}\mathbf{u}\\ {}^{\mathbf{2}}\mathbf{H}& \mathbf{2}\mathbf{.}\mathbf{014102}\mathbf{u}& {\mathbf{e}}^{\mathbf{\pm }}& \mathbf{0}\mathbf{.}\mathbf{0005486}\mathbf{u}\end{array} \\ {}^{\mathbf{3}}\mathit{H}\mathit{e}\mathbf{}\mathbf{}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{016029}\mathit{u} \end{gathered}$$

(Hint: Distinguish carefully between atomic and nuclear masses, and take the positrons properly into account.)

The expression for energy is given by,

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

Here, $\mathit{Q}$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.

Consider the first reaction.

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

The energy equation can be written as follows.

$Q_1=\left(2m_p-m_d-m_{e^{+}}\right)c^2$

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" $$\begin{gathered} Q_1=\left[2\left(m_{1H}-m_{e^{+}}\right)-\left(m_{2H}-m_{e^{+}}\right)-m_{e^{+}}\right]c^2 \\ =\left[2m_{1H}-m_{2H}-2m_{e^{+}}\right]c^2......\left(1\right) \end{gathered}$$

Substitute all the known values in equation (1).

$$\begin{gathered} Q_{1=}\left[2\left(1.007825u\right)-2.014102u-2\left(0.0005486u\right)\right]\left(931.5MeV/u\right) \\ =0.41984568MeV \\ \approx 0.42MeV \end{gathered}$$

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).

$$\begin{gathered} Q_2=\left[\left(2.014102u\right)+\left(1.007825u\right)-\left(3.016029u\right)\right]\left(931.5MeV/u\right) \\ =5.493987MeV \\ \approx 5.49MeV \end{gathered}$$

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).

$$\begin{gathered} Q_3=\left[2\left(3.016029u\right)-\left(4.002603u\right)-\left(1.007825u\right)\right]\left(931.5MeV/u\right) \\ =12.8593575MeV \\ \approx 12.86MeV \end{gathered}$$

Therefore, the energy released is $12.86MeV$.

Therefore, all the energy values are verified.