Chapter 30, Problem 47E

(a)

To determine

The energy released in the given process.

Answer to Problem 47E

The energy released in the process is $24.7\text{MeV}$.

Explanation of Solution

Write expression for mass defect.

    $\Delta m=4m_p-m_{\alpha }$        (1)

Here, $\Delta m$ is mass defect, $m_p$ is mass of proton and $m_{\alpha }$ is the mass of alpha particle.

Write expression for energy released.

    $E=\Delta m{c}^2$        (2)

Here, $c$ is speed of light.

Conclusion:

Substitute $1.00728\text{u}$ for $m_p$ and $4.0026\text{u}$ for $m_{\alpha }$ in equation (1).

    $\begin{array}{l}\Delta m=4\left(1.00728\text{u}\right)-4.0026\text{u}\\ \Delta m=0.02652\text{u}\end{array}$

Substitute $0.02652\text{u}$ for $\Delta m$ in equation (2).

    $\begin{array}{l}E=\left(0.02652\text{u}{c}^2\right)\left(\frac{931\text{MeV}}{1\text{u}{c}^2}\right)\\ E=24.7\text{MeV}\end{array}$

Thus, the energy released in the process is $24.7\text{MeV}$.

(b)

To determine

The mass of hydrogen burned.

Answer to Problem 47E

Themass of hydrogen burned is $1.99\times {10}^6\text{kg}$.

Explanation of Solution

Write expression for equivalent energy released by one hydrogen molecule.

    $E_{\text{eq}}=\frac{E{n}_{\alpha }}{n_h}$        (1)

Here, $n_{\alpha }$ is number of alpha particles formed and $n_h$ is number of hydrogen atoms.

Write expression for energy relareased by $1\text{kg}$ hydrogen.

    $E_H=E_{\text{eq}}\frac{N_A}{2}$        (2)

Here, $N_A$ is avogradro’s number.

Write expression for number of hydrogen atoms.

    $N=\frac{E_H}{E_{{\text{H}}_{\text{2}}\text{O}}}$        (3)

Here, $E_{{\text{H}}_{\text{2}}\text{O}}$ is energy released when one hydrogen molecule is used to produce water.

Write expression for mass of hydrogen.

    $m=\frac{2N}{N_A}$        (4)

Conclusion:

Substitute $24.7\text{MeV}$ for $E$, $1$ for $n_{\alpha }$ and $2$ for $n_h$ in equation (1).

    $\begin{array}{l}{E}_{\text{eq}}=\left(24.7\text{MeV}\right)\frac{1}{2}\\ E_{\text{eq}}=12.35\text{MeV}\end{array}$

Substitute $12.35\text{MeV}$ for $E_{\text{eq}}$ and $6.02\times {10}^{26}$ for $N_A$ in equation (2).

    $\begin{array}{c}{E}_H=12.35\text{MeV}\left(\frac{6.02\times {10}^{26}}{2}\right)\\ =3.7174\times {10}^{27}\text{MeV}\end{array}$

Substitute $3.7174\times {10}^{27}\text{MeV}$ for $E_H$ and $6.2\text{eV}$ for $E_{H_{2}O}$ in equation (3).

    $\begin{array}{l}N=\frac{3.7174\times {10}^{27}\text{MeV}}{6.2\text{eV}}\left(\frac{{10}^6\text{eV}}{1\text{MeV}}\right)\\ N=5.9958\times {10}^{32}\end{array}$

Substitute $5.9958\times {10}^{32}$ for $N$ and $6.02\times {10}^{26}$ for $N_A$ in equation (4).

    $\begin{array}{l}m=\frac{2\left(5.9958\times {10}^{32}\right)}{6.02\times {10}^{26}}\\ m=1.99\times {10}^6\text{kg}\end{array}$

Thus, mass of hydrogen burned is $1.99\times {10}^6\text{kg}$.