Question

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 at 0.4 m/s. The current density from anode to cthode is $4\mu A/m^2$. The velocity (in m/s) of positive ions moving towards cathode is

Answer 1

Step 1: Current density due to electrons

Charge carrier density for electrons,
$n=5\times {10}^{7}c{m}^{-3}$

$=5\times {10}^7\times {10}^6{m}^{-3}$

$=5\times {10}^{13}{m}^{-3}$

Drift velocity of electron, ($\nu d)=0.4m/s$

$e=1.6\times {10}^{-19}C$

We know, current density $J=ne{\nu }_d$

So, $J_e=5\times {10}^{13}{m}^{-3}\times 1.6\times {10}^{-19}C\times 0.4m/s$

$J_e=3.2\times {10}^{-6}A{m}^{-2}$

So, current density of drifing electron is $3.2\times {10}^{-6}A{m}^{-2}$

Step 2: Current density of positive ions

Total current density will be sum of the current due to electrons and positive ions as both will create current in same direction (anode to cathode)
total current density $J=4\times {10}^{-6}A{m}^{-2}$
So,current density for positive ions,

$J_p=J-J_e=4\times {10}^{-6}A{m}^{-2}-3.2\times {10}^{-6}A{m}^{-2}$

$J_p=0.8\times {10}^{-6}A{m}^{-2}$

Step 3: Drift speed of positive ions
$J_p=ne{v}_d$

$n=5\times {10}^{13}{m}^{-3}$

$(v_d{)}_p=\frac{J_p}{ne}$

$=\frac{0.8\times {10}^{-6}A{m}^{-2}}{5\times {10}^{13}{m}^{-3}\times 1.6\times {10}^{-19}C}=0.1m/s$

So, drift velocity of positive ions is 0.1 m/s.

Hence, velocity of positive ions moving towards cathode is 0.1 m/s.