Question

An ionisation chamber, with parallel conducting plates as anode and cathodes has $5\times 10^{7}c{m}^{-3}$ electrons and the same number of singly charged positive ions per $c{m}^3$. The electrons are moving toward the anode with velocity $0.4m{s}^{-1}$. The current density from anode to cathode is $4\mu A{m}^{-2}$. The velocity of positive ions moving towards cathode is

• A. $0.1m{s}^{-1}$
• B. $0.4m{s}^{-1}$
• C. zero
• D. $1.6m{s}^{-1}$

Answer 1

Answer: (A)

Here, number of electrons
$n_e=5\times 10^{7}c{m}^{-3}=5\times 10^7\times 10^6{m}^{-3}$
No. of positive ions,
$n_p=5\times 10^7\times 10^6=5\times 10^{13}{m}^{-3}$
$v=0.4m{s}^{-1};J=4\times 10^{-6}A{m}^{-2};v_p=?$
Use the relation
$J=n_{e}e{v}_e+n_{p}e{v}_p$ and solve it for $v_p$
$4\times 10^{-6}=\left(5\times 10^{13}\times 1.6\times 10^{-19}\times 0.4\right)$
$+\left(5\times 10^{13}\times 1.6\times 10^{-19}\times v_p\right)$
$v_p=\frac{4\times 10^{-6}-3.2\times 10^{-6}}{8.0\times 10^{-6}}$
$=\frac{0.8\times 10^{-6}}{8\times 10^{-6}}=0.1m{s}^{-1}$