Q. An ionisation chamber, with parallel conducting plates as anode and cathodes has $5 \times 10^{7}\ cm^{- 3}$ electrons and the same number of singly charged positive ions per $cm^{3}$. The electrons are moving toward the anode with velocity 0.4$ms^{- 1}$. The current density from anode to cathode is $4\mu Am^{- 2}.$ The velocity of positive ions moving towards cathode is
A) $0.1\ ms^{- 1}$
B) $0.4\ ms^{- 1}$
C) Zero
D) $1.6\ ms^{- 1}$
Explanation
(a) Here, No. of electrons
$n_{e} = 5 \times 10^{7}cm^{- 3} = 5 \times 10^{7} \times 10^{6}m^{- 3}$
No. of positive ions, $n_{p} = 5 \times 10^{7} \times 10^{6}$
$= 5 \times 10^{13}m^{- 3}$
$v = 0.4ms^{- 1};J = 4 \times 10^{- 6}\ Am^{- 2};v_{p} = ?$ Use the relation
$J = n_{e}e\ v_{e} + n_{p}\ e\ v_{p}$ and solve it for $v_{p}$
$4 \times 10^{- 6} = \left( 5 \times 10^{13} \times 1.6 \times 10^{- 19} \times 0.4 \right) + (5 \times 10^{13} \times 1.6 \times 10^{- 19} \times v_{p})$
$v_{p} = \frac{4 \times 10^{- 6} - 3.2 \times 10^{- 6}}{8.0 \times 10^{- 6}} = \frac{0.8 \times 10^{- 6}}{8 \times 10^{- 6}} = 0.1ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.