Question

An ionisation chamber, with parallel conducting plates as anode and cathode has singly charged positive $5 \times {10^7}$ ions per $c{m^3}$. The electrons are moving towards the anode with velocity of $0.4m/s$. The current density from anode to cathode is $4{\text{ mA/}}{{\text{m}}^2}$. The velocity of positive ions moving towards cathode is
(A) $0.1m/s$
(B) $0.4m/s$
(C) Zero
(D) $1.6m/s$

Answer 1

Hint
To solve this question, we have to use the formula of the current through a conductor in terms of the number density of the charge carriers. Then, on substituting the values of the quantities given in the question, we will get the final answer.
Formula Used: The formula used for solving this question is given by
$\Rightarrow i = neAv$, where $i$ is the current carried by the charge carrier having the number density $n$ and velocity $v$, through a conductor of area of cross section $A$.

Complete step by step answer
Let us consider the velocity of the electrons as ${v_e}$ and that of the positive ions as ${v_p}$
As a conductor is always neutral, so the number density of the electrons is equal to that of the positive ions.
As the ionization contains both the electrons and the positive ions, so both these will act as the charge carriers and therefore the total current will be the sum of the current due to both of them, that is
$\Rightarrow i = {i_e} + {i_p}$ (1)
We know that the current carried depends upon the number density of the charge carriers as
$\Rightarrow i = neAv$
So, the current carried by the electrons is given by the formula
$\Rightarrow {i_e} = neA{v_e}$ (2)
And the carried by the positive ions is given by the formula
$\Rightarrow {i_e} = neA{v_p}$ (3)
Substituting (2) and (3) in (1), we get
$\Rightarrow i = neA{v_e} + neA{v_p}$
Dividing both sides by $A$ we get
$\Rightarrow \dfrac{i}{A} = ne\left( {{v_e} + {v_p}} \right)$
We know that the current per unit area is called the current density $j$. So, we have
$\Rightarrow j = ne\left( {{v_e} + {v_p}} \right)$
Therefore, we get
$\Rightarrow {v_e} + {v_p} = \dfrac{j}{{ne}}$
So we get the velocity of the positive ions can be written as
$\Rightarrow {v_p} = \dfrac{j}{{ne}} - {v_e}$
According to the question, we have
$\Rightarrow n = 5 \times {10^7}/c{m^3}$
Converting it to meter we get,
$\Rightarrow n = \dfrac{{5 \times {{10}^7}}}{{{{10}^{ - 6}}}}/{m^3}$
On simplifying we get
$\Rightarrow n = 5 \times {10^{13}}/{m^3}$
Also, we have
$\Rightarrow j = 4{\text{ }}\mu {\text{A/}}{{\text{m}}^2} = 4 \times {10^{ - 6}}{\text{A/}}{{\text{m}}^2}$,
${v_e} = 0.4m/s$
And we know that the charge of an electron is given by $e = 1.6 \times {10^{ - 19}}C$. Substituting these values above, we get
$\Rightarrow {v_p} = \dfrac{{4 \times {{10}^{ - 6}}}}{{5 \times {{10}^{13}} \times 1.6 \times {{10}^{ - 19}}}} - 0.4$
On calculating this gives us,
$\Rightarrow {v_p} = 0.5 - 0.4$
Finally we get
$\Rightarrow {v_p} = 0.1m/s$
Thus the velocity of the positive ions comes out to be equal to $0.1m/s$.
Hence the correct answer is option (A).

Note
We should not forget to write all the quantities in their corresponding SI units. In this question, we can notice that the current density and the number density are not given in the SI units. So we are supposed to convert them to the corresponding SI unit before proceeding with the calculation.