Q. While doing his experiment, Millikan one day observed the following charges on a single drop (i)$\ 6.563 \times 10^{- 19}C\ $ (ii)$\ 8.204 \times 10^{- 19}C$ (iii)$11.50\ \times 10^{- 19}C\ $ (iv)$13.13 \times 10^{- 19}C$ (v)$\ 16.48 \times 10^{- 19}C\ \ $ (vi)$18.09 \times 10^{- 19}C$ From this data the value of the elementary charge $(e)$ was found to be
Explanation
(a) Any charge in the universe is given by
$q = ne \Rightarrow e = \frac{q}{n}\ $ (where$\ n\ $is an integer)
$q_{1}:q_{2}:q_{3}:q_{4}:q_{5}:q_{6} :: n_{1}:n_{2}:n_{3}:n_{4}:n_{5}:n_{6}$
$6.563\ :8.204\ :11.5\ :13.13\ :16.48\ :18.09$
$:: n_{1}:n_{2}\ :n_{3}:n_{4}:n_{5}:n_{6}$
Divide by $6.563$
$1\ :1.25\ :1.75\ :2.0\ :2.5\ :2.75\ ::n_{1}:n_{2}:n_{3}:n_{4}:n_{5}:n_{6}$
Multiplied by 4
$4\ :5\ :7\ :8\ :10\ :11\ ::\ n_{1}:n_{2}:n_{3}:n_{4}:n_{5}:n_{6}$
$e = \frac{q_{1} + q_{2} + q_{3} + q_{4} + q_{5} + q_{6}}{n_{1} + n_{2} + n_{3} + n_{4} + n_{5} + n_{6}} = \frac{73.967 \times 10^{- 19}}{45}$
$= 1.641 \times 10^{- 19}C$
[Note : If you take $45.0743$ in place of 45, you will get the exact value]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.