An ionization chamber with parallel conducting plates as anode and cathode has $5 \times 10^{7}$ electrons and the same number of singly-charged positive ions per $\mathrm{cm}^3$ . The electrons are moving at 0.4 m/s. The current density from anode to cathode is $4 \mu A / m^{2}$ . The velocity of positive ions moving towards cathode is
Text Solution
Verified by ExpertsD
Here, number of electrons
$n_{e} = 5 \times 10^{7} \text{ cm}^{-3} = 5 \times 10^{7} \times 10^{6} \text{ m}^{-3}$
No. of positive ions,
$n_{p} = 5 \times 10^{7} \times 10^{6} = 5 \times 10^{13} \mathrm{m}^{-3}$
$v = 0.4 \text{ ms}^{-1}; j = 4 \times 10^{-6} \text{ Am}^{-2}; v_p = ?$
Use the relation
$J = n_e e v_e + n_p e v_p$ and solve it for $v_p$
$4 \times 10^{-6} = (5 \times 10^{13} \times 1.6 \times 10^{-19} \times 0.4)$
$+(5 \times 10^{13} \times 1.6 \times 10^{-19} \times v_{p})$
$v_{p} = \frac{4 \times 10^{-6} - 3.2 \times 10^{-6}}{8.0 \times 10^{-6}} = \frac{0.8 \times 10^{-6}}{8 \times 10^{-6}} = 0.1 \, \mathrm{ms}^{-1}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems