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 towards the anode with velocity 0.4 m/s. The current density from anode to cathode is $4\mu A{{m}^{-2}}$. The velocity of positive ions moving towards cathode is
$\text{A}\text{. }0.4m{{s}^{-1}}$
$\text{B}\text{. zero}$
$\text{C}\text{. 1}.6m{{s}^{-1}}$
$\text{D}\text{. }0.1m{{s}^{-1}}$

Answer 1

Hint: The current between the anode and the cathode is equal to sum of currents due the electrons and the positively charged ions. Use the formula $i=neAv$ and find the currents due to both the charged particles. Current is equal to $\dfrac{i}{A}$. Using this, find the velocity of the positive ions.

Formula used:
 $i=neAv$

Complete step-by-step answer:
It is given that there are two conducting plates one as anode and the other as cathode. In an ionization chamber, an anode is a positively charged plate or plate that has higher potential. Therefore, the anode plate attracts the negatively charged particles, that are electrons.
On the other hand, cathode is a negatively charged plate or plate that has a lower potential. Therefore, the cathode plate attracts the positively charged ions.
Therefore, there will be flow of the electrons and the positive ions which will lead to a current. The current due singly charged particles is given as $i=neAv$, where n is the number of charged particles per unit volume, e is the charge of each particle, A is the area of cross section and v is the velocity of particles.
In this case, the total current will be the sum of the current due to the electrons (${{i}_{e}}$)and the positively charged ions (${{i}_{p}}$).
Here, n, e and A form both the current is the same. Therefore,
${{i}_{e}}=neA{{v}_{e}}$ and ${{i}_{p}}=neA{{v}_{p}}$.
Therefore, $i={{i}_{e}}+{{i}_{p}}=neA{{v}_{e}}+neA{{v}_{p}}$.
$\Rightarrow i=neA({{v}_{e}}+{{v}_{p}})$.
Current density is defined as the current per unit cross sectional area. Therefore, current density is equal to $\dfrac{i}{A}=ne({{v}_{e}}+{{v}_{p}})$ …. (i)
It is given that the current density is $4\mu A{{m}^{-2}}$.
It is also given that n = $5\times {{10}^{7}}c{{m}^{-3}}=5\times {{10}^{7}}\times {{\left( {{10}^{-2}}m \right)}^{-3}}=5\times {{10}^{3}}{{m}^{-3}}$.
e = $1.6\times {{10}^{-19}}C$ and ${{v}_{e}}=0.4m{{s}^{-1}}$.
Substitute the values in equation (i).
$\Rightarrow 4\mu =\left( 5\times {{10}^{13}} \right)\left( 1.6\times {{10}^{-19}} \right)(0.4+{{v}_{p}})$
$\Rightarrow \dfrac{\Rightarrow 4\times {{10}^{-6}}}{\left( 5\times {{10}^{13}} \right)\left( 1.6\times {{10}^{-19}} \right)}=(0.4+{{v}_{p}})$
$\Rightarrow (0.4+{{v}_{p}})=0.5$
$\Rightarrow {{v}_{p}}=0.1m{{s}^{-1}}$
This means that velocity of the positively charged ions is 0.1m/s.

So, the correct answer is “Option D”.

Note: Note that current is always measured in terms of positive charges. When electrons flow, we say the flow of electrons is the same as the flow of positrons with charge +e. Hence, we say that there is some current due the electrons in the opposite direction of their flow.