Question

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. $1.6\, ms ^{-1}$

Answer 1

Answer: (A)

Here, number of electrons
$n_{ e }=5 \times 10^{7} cm ^{-3}=5 \times 10^{7} \times 10^{6} m ^{-3}$
No. of positive ions,
$n_{p}=5 \times 10^{7} \times 10^{6}=5 \times 10^{13} m ^{-3}$
$v=0.4\, ms ^{-1} ;\, J =4 \times 10^{-6} Am ^{-2} ;\, v _{p}=?$
Use the relation
$J=n_{ e } ev _{ e }+n_{p} ev _{p}$ and solve it for $v_{p}$
$4 \times 10^{-6}=\left(5 \times 10^{13} \times 1.6 \times 10^{-19} \times 0.4\right)$
$+\left(5 \times 10^{13} \times 1.6 \times 10^{-19} \times v_{p}\right)$
$v_{p}= \frac{4 \times 10^{-6}-3.2 \times 10^{-6}}{8.0 \times 10^{-6}}$
$= \frac{0.8 \times 10^{-6}}{8 \times 10^{-6}}=0.1\, ms ^{-1}$