Question

The mobility of ${\mathbf{Na}}^{\mathbf{+}}$ions in water is$\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\left(m/s\right)\left(N/C\right)$. If an electric field of$\mathbf{2400}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{C}$is maintained in the fluid, what is the drift speed of the sodium ions?

Short answer

The drift speed of the sodium ions is $1.25\times {10}^{-4}\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Identification of the given data

The given data is listed below as,

• The mobility of the$N{a}^{+}$ ion in water is,$\mu =5.2\times {10}^{-8}\left(\mathrm{m}/\mathrm{s}\right)/\left(\mathrm{N}/\mathrm{C}\right)$
• The electric field maintained in the fluid is,$E=2500\mathrm{N}/\mathrm{C}$

Significance of the drift velocity

The drift velocity refers to the average velocity a charged particle attains due to the involvement of an electric field.

The equation of the drift velocity gives the drift speed of the sodium ions.

Determination of the drift speed of the sodium ion

The equation of the drift speed of the sodium ion is expressed as,

$v=\mu E$

Here, $\mu$is the mobility of the sodium ion and$E$is the electric field.

Substitute all the values in the above expression.

$$\begin{gathered} v=5.2\times {10}^{-8}\left(\mathrm{m}/\mathrm{s}\right)\left(\mathrm{N}/\mathrm{C}\right)\times 2400\mathrm{N}/\mathrm{C} \\ =1.25\times {10}^{-4}\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the drift speed of the sodium ions is $1.25\times {10}^{-4}\mathrm{m}/\mathrm{s}$.