Question

ILLUSTRATION 8 The resistivity of sea water is about 25 $\Omega$cm. The charge carriers are chiefly $Na^+$ and $Cl^-$ ions, and of each there are about 3x$10^{20}$ $cm^{-3}$. If we fill a plastic tube 2 m long with sea water and connect a 12 V battery to the electrodes at each end, what is the resulting average drift velocity of the ions, in $cms^{-1}$?

• A. 2.5 x $10^{-5}$ $cms^{-1}$
• B. 2.5 x $10^{-7}$ $cms^{-1}$
• C. 5.0 x $10^{-5}$ $cms^{-1}$
• D. 5.0 x $10^{-7}$ $cms^{-1}$

Answer 1

Answer: (A)

2.5 x $10^{-5}$ $cms^{-1}$

Answer 2

Explanation of the Solution

The problem asks for the average drift velocity of ions in sea water under a given electric field.

1. Identify given quantities and convert units:

   • Resistivity, $\rho = 25 \, \Omega \text{cm} = 0.25 \, \Omega \text{m}$.
   • Concentration of $Na^+$ ions, $n_{Na^+} = 3 \times 10^{20} \, \text{cm}^{-3}$.
   • Concentration of $Cl^-$ ions, $n_{Cl^-} = 3 \times 10^{20} \, \text{cm}^{-3}$.
   • Total charge carrier concentration, $n = n_{Na^+} + n_{Cl^-} = 6 \times 10^{20} \, \text{cm}^{-3}$.
   • Convert $n$ to SI units ($m^{-3}$): $n = 6 \times 10^{20} \times (10^2)^3 \, \text{m}^{-3} = 6 \times 10^{26} \, \text{m}^{-3}$.
   • Charge of an ion, $q = e = 1.6 \times 10^{-19} \, \text{C}$.
   • Length of the tube, $L = 2 \, \text{m}$.
   • Applied voltage, $V = 12 \, \text{V}$.

2. Calculate conductivity ($\sigma$) and electric field ($E$):

   • Conductivity is the reciprocal of resistivity: $\sigma = \frac{1}{\rho} = \frac{1}{0.25 \, \Omega \text{m}} = 4 \, \Omega^{-1} \text{m}^{-1}$.
   • Electric field strength: $E = \frac{V}{L} = \frac{12 \, \text{V}}{2 \, \text{m}} = 6 \, \text{V/m}$.

3. Relate current density ($J$), conductivity ($\sigma$), and electric field ($E$):

   • Using Ohm's Law in point form: $J = \sigma E$.
   • $J = (4 \, \Omega^{-1} \text{m}^{-1}) \times (6 \, \text{V/m}) = 24 \, \text{A/m}^2$.

4. Relate current density ($J$), carrier concentration ($n$), charge ($q$), and drift velocity ($v_d$):

   • The current density is also given by $J = nq v_d$, where $n$ is the total number of charge carriers per unit volume and $v_d$ is their average drift velocity.
   • Rearrange to solve for $v_d$: $v_d = \frac{J}{nq}$.

5. Calculate the drift velocity:

   • $v_d = \frac{24 \, \text{A/m}^2}{(6 \times 10^{26} \, \text{m}^{-3}) \times (1.6 \times 10^{-19} \, \text{C})}$.
   • $v_d = \frac{24}{9.6 \times 10^7} \, \frac{\text{A} \cdot \text{m}}{\text{C}} = 2.5 \times 10^{-7} \, \text{m/s}$.

6. Convert the drift velocity to the required units ($cms^{-1}$):

   • $v_d = 2.5 \times 10^{-7} \, \text{m/s} \times \frac{100 \, \text{cm}}{1 \, \text{m}} = 2.5 \times 10^{-5} \, \text{cm/s}$.