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 $c m^{3}$ . 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

0.4 m s^{- 1}

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z e r o

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1.6 m s^{- 1}

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0.1 m s^{- 1}

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Solution

The correct option is D $0.1 m s^{- 1}$
The current 'i' in the ionization chamber is given by,

$i = i_{e} + i_{p}$

where, $i_{e}$ is current due to electrons, $i_{p}$ is current due to positive ions

We can write $i = n e A v_{e} + n e A v_{p}$

Hence, $\frac{i}{A} = n e v_{e} + n e v_{p}$
where, A is area of cross section, $v_{e}$ is velocity of electrons and $v_{p}$ is velocity of positive ions

Hence, $4 \times 10^{- 6} = (5 \times 10^{13}) (1.6 \times 10^{- 19}) (v_{e} + v_{p})$

$0.5 = 0.4 + v_{p}$

$v_{p} = 0.1 m s^{- 1}$

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