Chapter 45, Problem 26P

(a)

To determine

Find the closest distance between the center of the nuclei.

Answer to Problem 26P

The closest distance between the center of the nuclei is $3.24\text{ fm}$.

Explanation of Solution

The deuterium-tritium fusion reaction,

    ${}_1^2\text{H}{+}_1^3\text{H}{\to }_2^4\text{He}{+}_0^1\text{n}$

Here, the tritium nucleus is at rest. The mass number of deuterium is $A_D=2$ and tritium is $A_T=3$.

Write the formula for radius of the nuclei

    $r=a{A}^{1/3}$                                                                             (I)

Where, $r$ is the radius of the nuclei, $a$ is a constant $\left(a=1.2\times {10}^{-15}\text{ m}\right)$ and $A$ is the mass number.

Conclusion:

The closest distance between the center of the two nuclei is $r=r_D+r_T$.

Substitute equation (I) in the above equation and solve

$r=a\left[A_D^{1/3}+A_T^{1/3}\right]$

Substitute $2$ for $A_D$, $3$ for $A_T$ and $1.20\times {10}^{-15}\text{ m}$ for $a$ in the above equation to find the value of $r$

$\begin{array}{l}r=1.20\times {10}^{-15}\text{ m}\times \left[{\left(2\right)}^{1/3}+{\left(3\right)}^{1/3}\right]\\ r=3.24\times {10}^{-15}\text{ m}\\ r=3.24\text{ fm}\end{array}$

Thus, the closest distance between the center of the nuclei is $3.24\text{ fm}$.

(b)

To determine

Find the electric potential energy at the closest distance between the center of the nuclei.

Answer to Problem 26P

The electric potential energy at the closest distance between the center of the nuclei is $444\text{ keV}$.

Explanation of Solution

The closest distance between the center of the nuclei is $3.24\text{ fm}$ and the atomic number of the reactant $Z_1=Z_2=1$. There is only one proton present in the nucleus of deuterium and tritium.

Write the formula for potential energy

  $U=k_e\frac{q_1{q}_2}{r}$                                                                                                 (II)

Where, $U$ is the potential energy, $k_e$ is the Boltzmann’s constant $\left(k_e=8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)$, $q_1\text{ and }{q}_2$ are the charges and $r$ is the distance between the two nuclei.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_e$, $e$ for $q_1$, $e$ for $q_2$ and $3.24\text{ fm}$ for $r$ in equation (II) to find the value of $U$. [Note: $e=1.60\times {10}^{-19}\text{ C}$]

$\begin{array}{l}U=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times {\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{3.24\times {10}^{-15}\text{ m}}\times \left(\frac{1\text{ keV}}{1.60\times {10}^{-16}\text{ J}}\right)\\ U=444\text{ keV}\end{array}$

Thus, the electric potential energy at the closest distance between the center of the nuclei is $444\text{ keV}$.

(c)

To determine

The speed of the deuterium and tritium nuclei as they touch.

Answer to Problem 26P

The speed of the deuterium and tritium nuclei as they touch is $v_{f=}\left(\frac{2}{5}\right)v_i$.

Explanation of Solution

The mass of deuterium is approximately $m_D=2\text{ u}$ and tritium is $m_T=3\text{ u}$.

According to the law of conservation of momentum,

  $\begin{array}{c}{m}_D{v}_i=\left(m_D+m_T\right)v_f\\ v_{f=}\left(\frac{m_D}{m_D+m_T}\right)v_i\end{array}$                                                                             (III)

Substitute $2\text{ u}$ for $m_D$ and $3\text{ u}$ for $m_T$ in the above equation to find the value of $v_f$

$\begin{array}{l}{v}_{f=}\left(\frac{2\text{ u}}{2\text{ u}+3\text{ u}}\right)v_i\\ v_{f=}\left(\frac{2}{5}\right)v_i\end{array}$

Thus, the speed of the deuterium and tritium nuclei as they touch is $v_{f=}\left(\frac{2}{5}\right)v_i$.

(d)

To determine

Find the minimum initial deuteron energy required to achieve fusion.

Answer to Problem 26P

The minimum initial deuteron energy required to achieve fusion is $740\text{ keV}$.

Explanation of Solution

According to the law of conservation of energy,

    $K_i+U_i=K_f+U_f$                                                                              (IV)

Here, $K_i$ is the initial kinetic energy of deuteron, $U_i$ is the initial potential energy of deuteron, $K_f$ is the final kinetic energy of deuteron and $U_f$ is the final potential energy of deuteron.

The deuteron has been moving from the beginning (infinity), therefore the initial potential energy of deuteron is zero, $U_i=0$.

Write the formula for kinetic energy

    $K=\frac{1}{2}m{v}^2$                                                                               (V)

Where, $K$ is the kinetic energy, $m$ is the mass and $v$ is the velocity.

Conclusion:

Substituting equation (V) in (IV),

$K_i+0=\frac{1}{2}\left(m_D+m_T\right)v_f^2+U_f$

Substitute (III) in the above equation,

$\begin{array}{l}{K}_i=\frac{1}{2}\left(m_D+m_T\right){\left(\frac{m_D}{m_D+m_T}\right)}^2{v}_i^2+U_f\\ K_i=\frac{1}{2}\left(m_D+m_T\right)\left(\frac{m_D}{m_D+m_T}\right)\times \left(\frac{m_D}{m_D+m_T}\right)v_i^2+U_f\\ K_i=\left(\frac{1}{2}{m}_D{v}_i^2\right)\left(\frac{m_D}{m_D+m_T}\right)+U_f\\ K_i=\left(\frac{m_D}{m_D+m_T}\right)K_i+U_f\end{array}$

$\begin{array}{c}{K}_i-\left(\frac{m_D}{m_D+m_T}\right)K_i=U_f\\ \left(1-\left(\frac{m_D}{m_D+m_T}\right)\right)K_i=U_f\\ \left(\frac{m_D+m_T-m_D}{m_D+m_T}\right)K_i=U_f\\ \left(\frac{m_T}{m_D+m_T}\right)K_i=U_f\end{array}$

Substitute $444\text{ keV}$ for $U_f$, $2\text{ u}$ for $m_D$ and $3\text{ u}$ for $m_T$ in the above equation to find the value of $K_i$

$\begin{array}{l}{K}_i=U_f\left(\frac{m_D+m_T}{m_T}\right)\\ K_i=444\text{ keV}\left(\frac{2\text{u}+3\text{u}}{3\text{u}}\right)\\ K_i=740\text{ keV}\end{array}$

Thus, the minimum initial deuteron energy required to achieve fusion is $740\text{ keV}$.

(e)

To determine

Why the fusion reaction occurs at much lower deuteron energies then the energy calculated in part (d).

Answer to Problem 26P

The fusion reaction occurs at much lower deuteron energies then the energy calculated must be possibly by tunneling through the potential energy barrier.

Explanation of Solution

Classically, the particle with energy $E<U$ should be reflected by the barrier with the height $U$. However, quantum mechanically the particles are able to tunnel through the barrier with lower energies and fuse together.

Therefore, the fusion reaction occurs at much lower deuteron energies then the energy calculated must be possibly by tunneling through the potential energy barrier.