Assume that a plasma temperature of$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{K}$ is reached in a laser-fusion device. (a) What is the most probable speed of a deuteron at that temperature? (b) How far would such a deuteron move in a confinement time of $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{s}$?

The Maxwell's speed distribution law is given by,

$\mathit{P}\left(v\right)\mathbf{=}\mathbf{4}\mathbf{\pi }{\left(\frac{M}{2\mathrm{\pi RT}}\right)}^{\mathbf{3}\mathbf{/}\mathbf{2}}{\mathbf{v}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}{\mathbf{Mv}}^{\mathbf{2}}\mathbf{/}\mathbf{2}\mathbf{RT}}$

(a)

To find the most probable speed, set the derivative of P(v) equals to zero, and then solve for v.

$$\begin{gathered} \frac{dP\left(v\right)}{dv}=0 \\ \frac{d}{dv}\left[v^2{e}^{-M{v}^2/2RT}\right]=0 \\ {v}^2\left(-\frac{2vM}{2RT}\right)e^{-M{v}^2/2RT}+2v{e}^{-M{v}^2/2RT}=0 \\ \frac{v^{2}M}{RT}=0 \\ v=\sqrt{\frac{2RT}{M}} \end{gathered}$$

But Nk=nR, for one mole n=1 and $N=N_A$,M is the molar mass and it equals the mass of a single particle multiplied by the number of particles in one mole $N_A$.

$$\begin{gathered} v_p=\sqrt{\frac{2RT}{M}} \\ =\sqrt{\frac{2RT}{m{N}_A}} \\ =\sqrt{\frac{2kt}{m}}.......\left(1\right) \end{gathered}$$

Substitute all the known values in equation (1).

$$\begin{gathered} v_p=\sqrt{\frac{2\left(1.38\times {10}^{-23}J/K\right)\left(1.0\times {10}^{8}K\right)}{3.343583\times {10}^{-27}kg}} \\ =9.1\times {10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the most probable speed of a deuteron is $9.1\times {10}^5\mathrm{m}/\mathrm{s}$.

(b)

The distance equals the time period multiplied by the velocity.

$r=v_p\Delta t\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}......\left(2\right)$

Substitute all the known values in equation (2).

$$\begin{gathered} r=\left(9.1\times {10}^5\mathrm{m}/\mathrm{s}\right)\left(1\times {10}^{-12}\mathrm{s}\right) \\ =9.1\times {10}^{-7}\mathrm{m} \end{gathered}$$

Therefore, the distance moved by deuteron is $9.1\times {10}^{-7}\mathrm{m}$.