Chapter 14, Problem 36P

To determine

The Q-value of the reactions in proton-proton cycle and the overall Q-value.

Answer to Problem 36P

The Q-value of the reactions in proton-proton cycle is $0.931\text{ MeV, 5}\text{.493 MeV, 19}\text{.285 MeV and 12}\text{.861 MeV}$ and the overall Q-value is $\text{12}\text{.861 MeV}$.

Explanation of Solution

Write the first reaction in proton-proton cycle

    $\text{H}+\text{H}\to \text{H}+e^{\text{+}}+\text{ν}$        (I)

Here, $\text{H}$ is the hydrogen, $\text{H}$ is the deuterium, $e^{+}$ is the electron and $\nu$ is the neutrino.

Write the expression for Q-value of this reaction

    $Q_1=\left[2m_{\left({}^{\text{1}}\text{H}\right)}-m_{\left({}^{\text{2}}\text{H}\right)}-m_{e^{+}}\right]c^2$        (II)

Here, $Q_1$ is the Q-value of the first reaction in the proton-proton cycle, $m_{\left({}^{\text{1}}\text{H}\right)}$ the mass of the hydrogen nucleus, $m_{\left({}^{\text{2}}\text{H}\right)}$ the mass of the deuterium, $m_{e^{+}}$ the mass of the positron and $c$ is the speed of light.

Write the second reaction in proton-proton cycle

    $\text{H}\text{+}\text{H}\to \text{H}\text{e }+\gamma$        (III)

Here, $\text{H}\text{e}$ is the Helium-3 nuclei and $\gamma$ is the gamma ray photon.

Write the expression for Q-value of this reaction

    $Q_2=\left[m_{\left({}^{\text{1}}\text{H}\right)}+m_{\left({}^{\text{2}}\text{H}\right)}-m_{\left({}^{\text{3}}\text{He}\right)}\right]c^2$        (IV)

Here, $Q_2$ is the Q-value of the second reaction in the proton-proton cycle and $m_{\left({}^{\text{3}}\text{He}\right)}$ the mass of the tritium.

Write the third reaction in proton-proton cycle

    $\text{H}+\text{H}\text{e}\to \text{H}\text{e}+e^{+}+\nu$        (V)

Here, $\text{H}\text{e}$ is the helium-4 nucleus.

Write the expression for Q-value of this reaction

    $Q_3=\left[m_{\left({}^{\text{1}}\text{H}\right)}+m_{\left({}^3\text{He}\right)}-m_{\left({}^4\text{He}\right)}-m_{e^{+}}\right]c^2$        (VI)

Here, $Q_3$ is the Q-value of the third reaction in the proton-proton cycle and $m_{\left({}^4\text{He}\right)}$ the mass of the helium-4 nuclei.

Write the fourth reaction in proton-proton cycle

    $\text{H}\text{e}+\text{H}\text{e}\to \text{H}\text{e}+\text{H}+\text{H}$        (VII)

Here, $\text{H}\text{e}$ is the helium-4 nucleus.

Write the expression for Q-value of this reaction

    $Q_4=\left[2m_{\left({}^3\text{He}\right)}-m_{\left({}^4\text{He}\right)}-2m_{{}^1\text{H}}\right]c^2$        (VIII)

Here, $Q_4$ is the Q-value of the fourth reaction in the proton-proton cycle.

Write the fourth reaction in proton-proton cycle

    $4\left(\text{H}\right)\to \text{H}\text{e}+{\text{2e}}^{+}+\text{2}\nu$        (IX)

Write the expression for Q-value of this reaction

    $Q=\left[4m_{{}^1\text{H}}-m_{\left({}^4\text{He}\right)}-2m_{e^{+}}\right]c^2$        (X)

Here, $Q$ is the Q-value of the overall reaction of the proton-proton cycle.

Conclusion:

Substitute $1.007825\text{u}$ for $m_{\left({}^{1}H\right)}$, $2.014102\text{u}$ for $m_{\left({}^{2}H\right)}$, $0.000549\text{u}$ for $m_{e^{+}}$ and $931.5\text{ Mev/u}$ for $c^2$ (II) to find $Q_1$

    $\begin{array}{c}{Q}_1=\left[2\left(1.007825\text{u}\right)-2.014102\text{u}-0.000549\text{u}\right]\left(931.5\text{ Mev/u}\right)\\ =0.931\text{ MeV}\end{array}$

Substitute $1.007825\text{u}$ for $m_{\left({}^{1}H\right)}$, $2.014102\text{u}$ for $m_{\left({}^{2}H\right)}$, $3.016029\text{u}$ for $m_{\left({}^3\text{He}\right)}$ and $931.5\text{ Mev/u}$ for $c^2$ (IV) to find $Q_2$

    $\begin{array}{c}{Q}_2=\left[\left(1.007825\text{u}\right)-2.014102\text{u}-3.016029\text{u}\right]\left(931.5\text{ Mev/u}\right)\\ =5.493\text{ MeV}\end{array}$

Substitute $1.007825\text{u}$ for $m_{\left({}^{1}H\right)}$, $3.016029\text{u}$ for $m_{\left({}^3\text{He}\right)}$, $4.002603\text{ u}$ for $m_{\left({}^4\text{He}\right)}$ $0.000549\text{u}$ for $m_{{\text{e}}^{-}}$ and $931.5\text{ Mev/u}$ for $c^2$ (VI) to find $Q_3$

    $\begin{array}{c}{Q}_3=\left[\left(1.007825\text{u}\right)+\left(3.016029\text{u}\right)-\left(4.002603\text{u}\right)-\left(0.000549\text{u}\right)\right]\left(931.5\text{ Mev/u}\right)\\ =19.285\text{ MeV}\end{array}$

Substitute $1.007825\text{u}$ for $m_{\left({}^{1}H\right)}$, $3.016029\text{u}$ for $m_{\left({}^3\text{He}\right)}$, $4.002603\text{ u}$ for $m_{\left({}^4\text{He}\right)}$ and $931.5\text{ Mev/u}$ for $c^2$ (VIII) to find $Q_4$

    $\begin{array}{c}{Q}_4=\left[2\left(3.016029\text{u}\right)-\left(4.002603\text{u}\right)-2\left(1.007825\text{u}\right)\right]\left(931.5\text{ Mev/u}\right)\\ =12.861\text{ MeV}\end{array}$

Substitute $1.007825\text{u}$ for $m_{\left({}^{1}H\right)}$, $4.002603\text{ u}$ for $m_{\left({}^4\text{He}\right)}$, $0.000549\text{u}$ for $m_{{\text{e}}^{-}}$ and $931.5\text{ Mev/u}$ for $c^2$ (X) to find $Q$

    $\begin{array}{c}Q=\left[4\left(1.007825\text{u}\right)-\left(4.002603\text{u}\right)-2\left(0.000549\text{u}\right)\right]\left(931.5\text{ Mev/u}\right)\\ =25.709\text{ MeV}\end{array}$

Therefore, the Q-value of the reactions in proton-proton cycle is $0.931\text{ MeV, 5}\text{.493 MeV, 19}\text{.285 MeV and 12}\text{.861 MeV}$ and the overall Q-value is $\text{12}\text{.861 MeV}$.