An ionization chamber with parallel conducting plates as anode and cathode has $5 x 10^{7}$ electrons and the same number of singly charged positive ions per cm. The electrons are moving towards the anode with velocity $0.4 m / s .$ The current density from anode to cathode is $4 µ A / m^{2} .$ The velocity of positive ions moving towards cathode is

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$0.4 m / s$

$1.6 m / s$

zero

$0.1 m / s$

answer is D.

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Detailed Solution

Current $= I_{c} + I_{p}$

$I_{c}$ and $I_{p}$ are current due to electrons and positively charged ions.

$\begin{array}{l} I = {neAV}_{d} \\ [n = 5 \times 10^{7} {cm}^{- 3} = 5 \times 10^{7} \times 10^{6} m^{- 3} = 5 \times 10^{13} m^{- 3}] \\ I_{e} = 5 \times 10^{13} \times 1.6 \times 10^{- 19} \times A \times 0.4 \\ I_{p} = 5 \times 10^{13} \times 1.6 \times 10^{19} \times A \times v \end{array}$

$\begin{array}{l} I = I_{c} + I_{p} \\ = 5 \times 10^{13} \times 1.6 \times 10^{- 19} \times A (v + 0.4) \end{array}$

Given, $I / A = 4 \times 10^{- 6} A / m^{2}$

$\begin{aligned} 4 \times 10^{- 6} \times A = 5 \times 10^{- 6} \times 1.6 \times A (v + 0.4) \\ \frac{4}{8} = v + 04 \Rightarrow 05 = v + 04 \Rightarrow v = 0.1 m / s . \end{aligned}$