Chapter 10, Problem 10.4P

(a)

To determine

The temperature required for the collision of two protons if quantum mechanical tunnelling is neglected.

Answer to Problem 10.4P

The temperature required for the collision of two protons is $1.1\times {10}^8\text{K}$.

Explanation of Solution

Write the expression for the reduced mass of a proton-proton collision by equating kinetic energy to thermal energy.

    $\frac{3}{2}kT=\frac{1}{2}{\mu }_p{v}_{\text{rms}}^2$

Here, $k$ is the Boltzmann constant, $T$ is the temperature and $v_{\text{rms}}$ is the root mean square velocity.

Conclusion:

Assume that nuclei having velocities ten times the root-mean square value for the Maxwell-Boltzmann distribution can overcome the Coulomb barrier. Therefore,

    $\begin{array}{c}v=10v_{\text{rms}}=10\times {\left(\frac{\left(3kT\right)}{{\mu }_p}\right)}^{1/2}\\ \frac{e^2}{4\pi {\epsilon }_{0}r}=\frac{1}{2}{\mu }_p{v}^2\\ \frac{e^2}{4\pi {\epsilon }_{0}r}=\frac{1}{2}\times 100\times \left(\frac{3}{2}kT\right)\end{array}$        (1)

Here, the separation between the charges to be $r\simeq 2\text{fm}$

Substitute $9\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}$ for $\frac{1}{4\pi {\epsilon }_0}$ in equation (1)

    $\begin{array}{l}T=\frac{4}{300k}\times \frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2\times \left(9\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}\right)}{2\times {10}^{-15}\text{m}}\\ T=1.1\times {10}^8\text{K}\end{array}$

Thus, the temperature required for the collision of two protons is $1.1\times {10}^8\text{K}$.

(b)

To determine

The ratio of the number of protons having velocities ten times the rms value to those moving at the rms velocity.

Answer to Problem 10.4P

The ratio of the number of protons having velocities ten times the rms value to those moving at the rms velocity is $5.7\times {10}^{31}$.

Explanation of Solution

Write the expression for ratio of atoms having velocities equal to ten times the rms value to those having the rms value.

    $\frac{n_v}{n_{\text{rms}}}=\frac{e^{-m{\left(10v_{\text{rms}}\right)}^{2/2kT}}{\left(10v_{\text{rms}}\right)}^2}{e^{-m{v}_{\text{rms}}^2/2kT}{v}_{\text{rms}}^2}$        (2)

Here, $n_v$ is the velocity of an atom and $n_{\text{rms}}$ is the value of root mean square.

Conclusion:

On solving equation (2)

    $\begin{array}{c}\frac{n_v}{n_{\text{rms}}}=100e^{-99\left(m{v}_{\text{rms}}^2/2kT\right)}\\ =100e^{-99\left(m/2kT\right)\left(3kT/2m\right)}\\ =100e^{-74.3}\\ =5.7\times {10}^{-31}\end{array}$

Thus,the ratio of the number of protons having velocities ten times the rms value to those moving at the rms velocity is $5.7\times {10}^{31}$.

(c)

To determine

The number of hydrogen nuclei in the sun.

Answer to Problem 10.4P

The number of hydrogen nuclei in the sun is $7\times {10}^{26}$.

Explanation of Solution

Write the expression for number of atomic nuclei in the sun.

    $N_H=N\left(\frac{n_v}{n_{\text{rms}}}\right)$        (3)

Here, $N_H$ is the number of atomic nuclei in the sun.

Conclusion:

Substitute $5.7\times {10}^{31}$ for $\left(\frac{n_v}{n_{\text{rms}}}\right)$ and $1.2\times {10}^{57}$ for $N$ in equation (3).

    $\begin{array}{c}{N}_H=1.2\times {10}^{57}\times 5.7\times {10}^{31}\\ =7\times {10}^{26}\end{array}$

Thus,the number of hydrogen nuclei in the sun is $7\times {10}^{26}$.