Question

An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 \times {10^4}{\text{N}}{{\text{C}}^{{\text{ - 1}}}}$ in Millikan's oil drop experiment. The density of the oil is $1.26{\text{gc}}{{\text{m}}^{{\text{ - 3}}}}$ . Estimate the radius of the drop. $\left[ {{\text{g}} = 9.8{\text{m}}{{\text{s}}^{{\text{ - 2}}}};{\text{e}} = 1.60 \times {{10}^{ - 19}}{\text{C}}} \right]$ .

Answer 1

Hint: Millikan’s oil drop experiment was performed to measure the charge of an electron. Before this experiment, the existence of subatomic particles was not accepted universally.
The apparatus used in Millikan’s oil drop experiment, the apparatus used contained an electric field created between a pair of parallel metal plates which were held by an insulating material. The droplets which are electrically charged entered the electric field and were balanced between two plates by altering the field.

Complete step by step answer:
In this experiment, a negatively charged oil drop is suspended in a uniform electric field such that its weight is balanced by the electric force of the field. The net force on the charge is zero.
Thus,
 $
  {\text{mg}} = {\text{qE}} \\
   \Rightarrow {\text{m}} = \dfrac{{{\text{qE}}}}{{\text{g}}} \\
 $
Here, E represents the electric field force, q is the charge of the electron, m is the weight of the oil drop and g represents the gravity.
Given that there are 12 excess electrons, so the charge q will be 12 times the charge of one electron. So, ${\text{q}} = 12 \times 1.6 \times {10^{ - 19}}{\text{C}}$ .
Also given that E is equal to $2.55 \times {10^4}{\text{N}}{{\text{C}}^{{\text{ - 1}}}}$ .
Therefore, substituting these in the above expression, we have:
$
  {\text{m}} = \dfrac{{12 \times 1.6 \times {{10}^{ - 19}} \times 2.55 \times {{10}^4}}}{{9.81}} \\
   \Rightarrow {\text{m}} = 4.99 \times {10^{ - 15}}{\text{kg}} \\
 $
The density of the oil, ${{\rho }}$ is given to be $1.26{\text{gc}}{{\text{m}}^{{\text{ - 3}}}}$ or $1.26 \times {10^3}{\text{kgc}}{{\text{m}}^{{\text{ - 3}}}}$
We know, volume V is related to the density by the relation:
${\text{V}} = \dfrac{{\text{m}}}{{{\rho }}}$
Therefore, the volume of the oil drop is:
\[
  {\text{V}} = \dfrac{{4.99 \times {{10}^{ - 15}}}}{{1.26 \times {{10}^3}}} \\
   \Rightarrow {\text{V}} = 3.96 \times {10^{ - 18}}{{\text{m}}^{\text{3}}} \\
 \]
If the drop is assumed to be spherical, then volume is related to the radius of the drop r as:
${\text{V}} = \dfrac{4}{3}{{\pi }}{{\text{r}}^{\text{3}}}$
Substitute all the values,
\[
  {{\text{r}}^{\text{3}}} = \dfrac{{3 \times 3.96 \times {{10}^{ - 18}}}}{{4 \times 3.14}} \\
   \Rightarrow {{\text{r}}^{\text{3}}} = 0.9458 \times {10^{ - 18}} \\
   \Rightarrow {\text{r}} = 0.98 \times {10^{ - 6}}{\text{m}} \\
 \]

Note:
The conclusion of this oil drop experiment is that the charge over any oil droplet is always an integral value of $1.6 \times {10^{ - 19}}$ . Or in other words, the charge is said to be quantized, i.e., the charge on any particle will always be an integral multiple of $1.6 \times {10^{ - 19}}$ .