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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
$1.641 \times {10^{ - 19}}C$
$1.630 \times {10^{ - 19}}C$
$1.648 \times {10^{ - 19}}C$
$1.602 \times {10^{ - 19}}C$
Solution
(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)
