An ionization chamber with parallel conducting plates as anode and cathode, has 5 x 107 electrons and the same number of singly charged positive ions per cm3 . The electrons are moving towards the anode with velocity 0.4 m/s. The current density from anode to cathode is 4μ A/m 2 . The velocity of positive ions moving towards cathode is

i=ie+ip=neAV1+V2;1A=neV1+V24×10−6=5×10710−61.6×10−190.4+V20.5=0.4+V2;V2=0.1m/s

An ionization chamber with parallel conducting plates as anode and cathode, has 5 x 107 electrons and the same number of singly charged positive ions per cm3 . The electrons are moving towards the anode with velocity 0.4 m/s. The current density from anode to cathode is 4μ A/m 2 . The velocity of positive ions moving towards cathode is

i=ie+ip=neAV1+V2;1A=neV1+V24×10−6=5×10710−61.6×10−190.4+V20.5=0.4+V2;V2=0.1m/s

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An ionization chamber with parallel conducting plates as anode and cathode, has 5 x 10⁷ electrons and the same number of singly charged positive ions per cm³ . 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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zero

1.6 ms^-1

0.4ms^-1

0.1 ms^-1

answer is D.

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Detailed Solution

$\begin{array}{l} i = (i_{e} + i_{p}) = neA (V_{1} + V_{2}); \frac{1}{A} = ne (V_{1} + V_{2}) \\ 4 \times 10^{- 6} = \frac{5 \times 10^{7}}{10^{- 6}} 1 .6 \times 10^{- 19} (0 .4 + V_{2}) \\ 0 .5 = 0 .4 + V_{2}; V_{2} = 0 .1 m / s \end{array}$

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